Quadratic Equations · Class 10 — 170 MCQs with Answers

Q1. The standard form of a quadratic equation is

Beginner

A $ax^2 + bx + c = 0,\ a = 0$ B $ax + b = 0$ C $ax^3 + bx^2 + c = 0$ D $ax^2 + bx + c = 0,\ a \ne 0$

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$ax^2 + bx + c = 0,\ a \ne 0$

Degree two requires a non-zero coefficient of $x^2$.

Q2. Setting $a = 0$ in $ax^2 + bx + c = 0$ turns it into

Beginner

A a cubic equation B a quadratic equation still C no equation at all D a linear equation

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a linear equation

Without the $x^2$ term only $bx + c = 0$ is left, which has degree one.

Q3. Which of these is a quadratic equation?

Beginner

A $3x - 7 = 0$ B $x^3 + 1 = 0$ C $\frac{1}{x} = 4$ D $x^2 - 9 = 0$

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$x^2 - 9 = 0$

Only the first has highest power 2 with a non-zero coefficient.

Q4. In $x^2 - 5x + 6 = 0$, the value of b is

Beginner

A 5 B 6 C 1 D −5

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−5

The coefficient of x carries its sign.

Q5. In $3x^2 - 7 = 0$, the value of b is

Beginner

A 3 B −7 C 7 D 0

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0

There is no x term, so its coefficient is 0.

Q6. In $2x^2 + x - 528 = 0$, the value of c is

Beginner

A 528 B −528 C 2 D 1

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−528

The constant term, with its sign.

Q7. A quadratic equation has at most how many roots?

Beginner

A 2 B 1 C 3 D infinitely many

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2

Degree two allows at most two roots.

Q8. A number $\alpha$ is a root of $ax^2 + bx + c = 0$ when

Beginner

A $a\alpha^2 + b\alpha + c = 1$ B $a\alpha^2 + b\alpha + c = 0$ C $\alpha = 0$ D $\alpha = a$

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$a\alpha^2 + b\alpha + c = 0$

Substituting a root makes the left side vanish.

Q9. Is 1 a root of $2x^2 - 3x + 1 = 0$?

Beginner

A yes, because $2 - 3 + 1 = 0$ B no, because it gives 1 C no, because it gives 6 D only if x is replaced by −1

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yes, because $2 - 3 + 1 = 0$

Substituting $x = 1$ gives exactly zero.

Q10. Two consecutive positive integers are written as

Beginner

A x and $x + 2$ B x and $2x$ C x and $x - 2$ D x and $x + 1$

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x and $x + 1$

Consecutive integers differ by 1.

Q11. Two consecutive EVEN integers are written as

Beginner

A x and $x + 1$ B x and $2x$ C x and $x + 2$ D $2x$ and $2x + 1$

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x and $x + 2$

Even numbers are two apart.

Q12. Assertion (A): $x^2 - 9 = 0$ is a quadratic equation although it has no x term. Reason (R): The standard form permits $b = 0$; only $a$ must be non-zero.

Beginner

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

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Both A and R are true, and R is the correct explanation of A

A missing middle term simply means $b = 0$, which the definition allows.

Q13. Which equation is still quadratic even though its constant term is zero?

Beginner

A $3x + 5 = 0$ B $\frac{3}{x} + 5 = 0$ C $3x^2 + 5x = 0$ D $3x^3 = 5$

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$3x^2 + 5x = 0$

Only a is forbidden from being zero, so $c = 0$ leaves the equation quadratic.

Q14. In $7x^2 = 3$, written in standard form, the value of c is

Beginner

A 3 B 7 C 0 D −3

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−3

Standard form is $7x^2 - 3 = 0$, so the constant is $-3$.

Q15. The degree of a quadratic equation is

Beginner

A 1 B 2 C 3 D 0

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2

The highest power of the variable is 2.

Q16. Is 3 a root of $x^2 - x - 6 = 0$?

Beginner

A no, because it gives 6 B no, because it gives 12 C yes, because $9 - 3 - 6 = 0$ D only if x is replaced by −3

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yes, because $9 - 3 - 6 = 0$

Substituting $x = 3$ gives zero, so 3 is a root.

Q17. Assertion (A): In $x^2 - 7 = 0$ the coefficient b equals 0. Reason (R): The equation contains no term in x, so its coefficient is zero.

Beginner

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

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Both A and R are true, and R is the correct explanation of A

A missing term is a term with coefficient zero, which is exactly what R says.

Q18. Check whether $(x + 1)^2 = 2(x - 3)$ is a quadratic equation.

Beginner

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Q19. Show that 1 is a root of $2x^2 - 3x + 1 = 0$, and that 2 is not.

Beginner

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Q20. The product of two consecutive positive integers is 306. Represent this as a quadratic equation.

Beginner

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Q21. The zero-product rule states that if $pq = 0$ then

Beginner

A $p = 0$ or $q = 0$ B $p = 0$ and $q = 0$ C $p = q$ D $p + q = 0$

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$p = 0$ or $q = 0$

At least one factor must vanish; both may, but only one is required.

Q22. The roots of $(x - 4)(x + 5) = 0$ are

Beginner

A −4 and 5 B 4 and 5 C 4 and −5 D −4 and −5

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4 and −5

Each bracket is set to zero, and the sign flips each time.

Q23. To factorise $ax^2 + bx + c$, the two numbers must multiply to

Beginner

A $ac$ B $a + c$ C $b$ D $a$

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$ac$

Their product is $ac$ and their sum is $b$.

Q24. The sum of the two numbers used to split the middle term equals

Beginner

A $ac$ B $a + c$ C $c$ D $b$

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$b$

Their sum is the coefficient of the middle term.

Q25. The roots of $x^2 - 5x + 6 = 0$ are

Beginner

A −2 and −3 B 2 and 3 C 1 and 6 D −1 and −6

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2 and 3

$(x-2)(x-3) = 0$.

Q26. Factorising $x^2 - 9 = 0$ gives the roots

Beginner

A 9 and −9 B 3 and −3 C 3 only D 0 and 9

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3 and −3

$(x-3)(x+3) = 0$.

Q27. The roots of $3x^2 + 5x = 0$ are

Beginner

A 0 and $\frac53$ B 0 and $-\frac53$ C $-\frac53$ only D 3 and 5

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0 and $-\frac53$

$x(3x + 5) = 0$, so $x = 0$ or $3x = -5$.

Q28. How many roots does $(x + 7)^2 = 0$ have?

Beginner

A one root B two distinct roots C no roots D two equal roots

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two equal roots

Both factors give $x = -7$, so the roots are equal.

Q29. Before factorising, an equation must be written

Beginner

A with the largest term last B as a product C in standard form with 0 on one side D with positive coefficients only

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in standard form with 0 on one side

The zero-product rule needs a zero on one side.

Q30. The factor $(2x - 3)$ gives the root

Beginner

A $-\frac32$ B $\frac32$ C 3 D $\frac23$

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$\frac32$

$2x - 3 = 0$ gives $x = \frac32$.

Q31. Assertion (A): From $(x - 1)(x - 2) = 6$ you may write $x - 1 = 6$. Reason (R): The zero-product rule applies only when the product equals zero.

Beginner

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

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A is false but R is true

A is false — the rule cannot be used against 6 — and R is the true statement that explains why.

Q32. Solving $x^2 + 7x + 12 = 0$ gives

Beginner

A 3 and 4 B −2 and −6 C −3 and −4 D 2 and 6

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−3 and −4

$ac = 12$, $b = 7$, giving 3 and 4, so $(x+3)(x+4) = 0$.

Q33. Which pair of numbers solves $x^2 - 2x - 8 = 0$?

Beginner

A −4 and 2 B 8 and −1 C 1 and −8 D 4 and −2

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4 and −2

$ac = -8$, $b = -2$, giving $-4$ and 2, so $(x-4)(x+2) = 0$.

Q34. Assertion (A): $x^2 - 9 = 0$ can be solved by factorisation. Reason (R): It is a difference of two squares, $(x-3)(x+3)$.

Beginner

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

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Both A and R are true, and R is the correct explanation of A

The difference of squares is a factorisation, so R is precisely why A holds.

Q35. Solve by factorisation: $x^2 - 3x - 10 = 0$.

Beginner

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Q36. Solve by factorisation: $2x^2 + x - 6 = 0$.

Beginner

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Q37. The discriminant of $ax^2 + bx + c = 0$ is

Beginner

A $b^2 + 4ac$ B $4ac - b^2$ C $b^2 - 4ac$ D $\sqrt{b^2 - 4ac}$

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$b^2 - 4ac$

It is the expression under the square root in the formula.

Q38. If $D > 0$, the equation has

Beginner

A two distinct real roots B two equal roots C no real roots D exactly one root

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two distinct real roots

A positive discriminant gives two different values from the $\pm$.

Q39. A zero discriminant tells you that the curve $y = ax^2 + bx + c$

Beginner

A touches the x-axis at exactly one point B cuts the x-axis twice C never meets the x-axis D lies along the x-axis

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touches the x-axis at exactly one point

Equal roots mean the two crossing points have merged into a single point of contact.

Q40. Which discriminant value means the equation has no solution among the real numbers?

Beginner

A 9 B $-9$ C 0 D 1

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$-9$

Only a negative discriminant leaves a negative quantity under the square root.

Q41. The quadratic formula is

Beginner

A $x = \frac{b \pm \sqrt{b^2 - 4ac}}{2a}$ B $x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2a}$ C $x = -b \pm \frac{\sqrt{b^2 - 4ac}}{2a}$ D $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

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$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The numerator is $-b \pm \sqrt{D}$ and the whole of it is divided by $2a$.

Q42. For $x^2 - 4x + 3 = 0$, the discriminant is

Beginner

A −4 B 28 C 4 D 16

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4

$16 - 12 = 4$.

Q43. For $x^2 - 4x + 4 = 0$, the discriminant is

Beginner

A 4 B 0 C −4 D 32

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0

$16 - 16 = 0$, so the roots are equal.

Q44. For $x^2 - 4x + 6 = 0$, the discriminant is

Beginner

A 8 B 40 C −40 D −8

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−8

$16 - 24 = -8$, so there are no real roots.

Q45. If $b = -7$, then $b^2$ equals

Beginner

A −49 B 14 C 49 D −14

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49

Squaring a negative gives a positive.

Q46. If $b = -7$, then $-b$ in the formula equals

Beginner

A −7 B 7 C 49 D $\frac17$

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7

The sign flips when b is negated.

Q47. The quadratic formula may be used when

Beginner

A $b^2 - 4ac < 0$ B $b^2 - 4ac \ge 0$ C $a = 0$ D $c = 0$

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$b^2 - 4ac \ge 0$

A real square root needs a non-negative quantity.

Q48. Assertion (A): $x^2 + 4 = 0$ has two distinct real roots. Reason (R): Its discriminant is $0 - 16 = -16$, which is negative.

Beginner

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

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A is false but R is true

A negative discriminant leaves no real root at all, so the assertion is false. The reason computes that discriminant correctly.

Q49. In the formula, the expression under the square root sign is

Beginner

A the coefficient a B twice the constant C the sum of the roots D the discriminant

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the discriminant

$b^2 - 4ac$ sits under the root, which is why its sign decides everything.

Q50. For $x^2 - 8x + 16 = 0$, the discriminant is

Beginner

A 0 B 64 C −64 D 128

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0

$64 - 64 = 0$.

Q51. When $D = 0$, both roots are equal to

Beginner

A $-\frac{b}{2a}$ B $\frac{b}{2a}$ C $-\frac{c}{a}$ D $\frac{-b}{a}$

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$-\frac{b}{2a}$

The $\pm\sqrt{0}$ contributes nothing, leaving $\frac{-b}{2a}$ twice.

Q52. Find the discriminant of $2x^2 - 4x + 3 = 0$ and state the nature of its roots.

Beginner

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Q53. Find the discriminant of $3x^2 - 2x - 1 = 0$ and state the nature of its roots.

Beginner

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Q54. Without solving, state how many real roots $x^2 - 4x + 4 = 0$ has, and explain.

Beginner

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Q55. Is $(x - 2)(x + 1) = (x - 1)(x + 3)$ a quadratic equation?

Intermediate

A yes, with $a = 1$ B no — the $x^2$ terms cancel and it is linear C yes, with $a = 2$ D no — it is cubic

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no — the $x^2$ terms cancel and it is linear

Expanding gives $x^2 - x - 2 = x^2 + 2x - 3$, so $3x = 1$.

Q56. Simplifying $x^2 + 3x + 1 = (x - 2)^2$ leaves an equation of degree

Intermediate

A 2 B 1 C 3 D 0

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1

$(x-2)^2 = x^2 - 4x + 4$, so the $x^2$ terms cancel and $7x = 3$ remains — degree one.

Q57. Expanding $(x + 1)^2 = 2(x - 3)$ and collecting gives

Intermediate

A $x^2 - 7 = 0$ B $x^2 + 4x + 7 = 0$ C $x^2 + 7 = 0$ D $2x + 7 = 0$

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$x^2 + 7 = 0$

$x^2 + 2x + 1 = 2x - 6$; the x terms cancel, leaving $x^2 + 7 = 0$ — still degree two.

Q58. Is $(x + 2)^3 = 2x(x^2 - 1)$ a quadratic equation?

Intermediate

A yes B no — it is linear C no — it is cubic D no — it is of degree 4

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no — it is cubic

Expanding leaves $-x^3 + 6x^2 + 14x + 8 = 0$, of degree 3.

Q59. Simplifying $x^3 - 4x^2 - x + 1 = (x - 2)^3$ gives

Intermediate

A a cubic equation B $2x^2 + 13x + 9 = 0$ C $2x^2 - 13x + 9 = 0$ D $x^2 - 13x + 9 = 0$

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$2x^2 - 13x + 9 = 0$

$(x-2)^3 = x^3 - 6x^2 + 12x - 8$; the $x^3$ terms cancel and $2x^2 - 13x + 9 = 0$ remains.

Q60. Writing $x^2 - 2x = -2(3 - x)$ in standard form gives

Intermediate

A $x^2 - 4x + 6 = 0$ B $x^2 - 2x + 6 = 0$ C $x^2 + 6 = 0$ D $x^2 - 4x - 6 = 0$

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$x^2 - 4x + 6 = 0$

The right side is $-6 + 2x$, so $x^2 - 2x - 2x + 6 = 0$.

Q61. A rectangular plot has area 528 m² and its length is one metre more than twice its breadth. Taking the breadth as x, the equation is

Intermediate

A $2x^2 + x - 528 = 0$ B $x^2 + 2x - 528 = 0$ C $2x^2 - x - 528 = 0$ D $2x^2 + x + 528 = 0$

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$2x^2 + x - 528 = 0$

Length $2x + 1$, so $x(2x + 1) = 528$.

Q62. The product of two consecutive positive integers is 306. The equation is

Intermediate

A $x^2 + 2x - 306 = 0$ B $x^2 - x - 306 = 0$ C $x^2 + x - 306 = 0$ D $2x^2 - 306 = 0$

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$x^2 + x - 306 = 0$

$x(x + 1) = 306$.

Q63. Rohan's mother is 26 years older than him, and in 3 years the product of their ages will be 360. Taking Rohan's present age as x, the equation is

Intermediate

A $x^2 + 26x - 360 = 0$ B $x^2 + 32x - 273 = 0$ C $x^2 + 29x - 360 = 0$ D $x^2 + 32x + 273 = 0$

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$x^2 + 32x - 273 = 0$

In 3 years the ages are $x + 3$ and $x + 29$, so $(x+3)(x+29) = 360$.

Q64. The sum of the squares of two consecutive positive integers is 365. The equation is

Intermediate

A $x^2 + x - 365 = 0$ B $2x^2 - 365 = 0$ C $x^2 + x - 182 = 0$ D $x^2 + 2x - 365 = 0$

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$x^2 + x - 182 = 0$

$x^2 + (x+1)^2 = 365$ gives $2x^2 + 2x - 364 = 0$; dividing by 2 gives $x^2 + x - 182 = 0$.

Q65. Assertion (A): An equation containing $x^2$ is always quadratic. Reason (R): The degree of an equation is found after expanding and simplifying it.

Intermediate

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

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A is false but R is true

A is false — the $x^2$ terms can cancel — while R is true and is exactly why A fails.

Q66. Assertion (A): The roots of $x^2 - 5x + 6 = 0$ are the zeroes of the polynomial $x^2 - 5x + 6$. Reason (R): A root of an equation makes its left-hand side zero, and a zero of a polynomial makes the polynomial zero.

Intermediate

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

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Both A and R are true, and R is the correct explanation of A

The two definitions are the same condition, so they pick out the same numbers.

Q67. Writing $4x(x - 3) = 2(x + 5)$ in standard form gives

Intermediate

A $4x^2 - 10x - 10 = 0$ B $4x^2 - 14x - 10 = 0$ C $4x^2 - 14x + 10 = 0$ D $4x^2 - 12x - 10 = 0$

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$4x^2 - 14x - 10 = 0$

$4x^2 - 12x = 2x + 10$, so $4x^2 - 14x - 10 = 0$.

Q68. A garden is 6 m longer than it is wide and covers 187 m². Taking the width as x, the equation is

Intermediate

A $x^2 - 6x - 187 = 0$ B $x^2 + 6x + 187 = 0$ C $x^2 + 6x - 187 = 0$ D $2x^2 + 6x - 187 = 0$

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$x^2 + 6x - 187 = 0$

The length is $x + 6$, so $x(x + 6) = 187$.

Q69. The product of two consecutive EVEN positive integers is 288. The equation is

Intermediate

A $x^2 + x - 288 = 0$ B $x^2 + 2x - 144 = 0$ C $2x^2 + x - 288 = 0$ D $x^2 + 2x - 288 = 0$

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$x^2 + 2x - 288 = 0$

Even integers differ by 2, so $x(x + 2) = 288$.

Q70. Check whether $(x - 2)(x + 1) = (x - 1)(x + 3)$ is a quadratic equation.

Intermediate

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Q71. Check whether $(x + 2)^3 = 2x(x^2 - 1)$ is a quadratic equation.

Intermediate

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Q72. Write $(x - 3)(2x + 1) = x(x + 5)$ in standard form and state a, b and c.

Intermediate

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Q73. The area of a rectangular plot is 528 m² and its length is one metre more than twice its breadth. Represent this as a quadratic equation.

Intermediate

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Q74. Explain why an equation containing $x^2$ need not be quadratic, with an example.

Intermediate

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Q75. If 4 is a root of $x^2 - 9x + m = 0$, find m.

Intermediate

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Q76. (a) State the standard form of a quadratic equation and explain why $a \ne 0$. (b) Check whether each of these is quadratic, showing your working: (i) $(2x - 1)(x - 3) = (x + 5)(x - 1)$, (ii) $x^2 + 3x + 1 = (x - 2)^2$. (c) For the one that is quadratic, state a, b and c.

Intermediate

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1½ marks — (a) the form and the reason. 2½ marks — (b) both expansions and both verdicts. 1 mark — (c) a, b and c with correct signs.

Q77. (a) Explain what it means for a number to be a root of a quadratic equation. (b) Show that $-2$ is a root of $x^2 - x - 6 = 0$ but 2 is not. (c) Test whether 4 is a root of the same equation, and state how many roots a quadratic equation can have.

Intermediate

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1 mark — (a) the definition. 2 marks — (b) both substitutions. 2 marks — (c) the third substitution and the statement about the number of roots.

Q78. The correct split of the middle term of $2x^2 - 5x + 3$ is

Intermediate

A $-1x - 4x$ B $-2x + 3x$ C $2x - 7x$ D $-2x - 3x$

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$-2x - 3x$

$ac = 6$ and $b = -5$, so the numbers are $-2$ and $-3$.

Q79. The roots of $2x^2 - 5x + 3 = 0$ are

Intermediate

A $-1$ and $-\frac32$ B 2 and 3 C $\frac12$ and 3 D 1 and $\frac32$

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1 and $\frac32$

$(x-1)(2x-3) = 0$.

Q80. The roots of $6x^2 - x - 2 = 0$ are

Intermediate

A $-\frac23$ and $\frac12$ B 2 and $-1$ C $\frac23$ and $-\frac12$ D $\frac13$ and $-\frac12$

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$\frac23$ and $-\frac12$

$(3x-2)(2x+1) = 0$.

Q81. The roots of $100x^2 - 20x + 1 = 0$ are

Intermediate

A $\frac{1}{10}$ and $-\frac{1}{10}$ B 10 and 10 C $\frac{1}{10}$ and $\frac{1}{10}$ D $\frac{1}{20}$ and $\frac{1}{5}$

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$\frac{1}{10}$ and $\frac{1}{10}$

$(10x - 1)^2 = 0$, so the roots are equal.

Q82. After grouping, the two brackets come out different. This means

Intermediate

A the split was wrong B the equation has no roots C the roots are equal D the equation is not quadratic

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the split was wrong

A correct split always leaves the same bracket in both pairs.

Q83. Dividing $2x^2 - 14x - 120 = 0$ by 2 first gives

Intermediate

A $x^2 - 14x - 60 = 0$ B $x^2 - 7x - 60 = 0$ C $x^2 - 7x - 120 = 0$ D $2x^2 - 7x - 60 = 0$

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$x^2 - 7x - 60 = 0$

Every term is divided, including the constant.

Q84. Solving $x^2 - 7x - 60 = 0$ by factorisation gives

Intermediate

A 12 and −5 B −12 and 5 C 10 and −6 D 15 and −4

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12 and −5

$ac = -60$ and $b = -7$, giving $-12$ and 5, so $(x-12)(x+5) = 0$.

Q85. Splitting the middle term of $x^2 + x - 306 = 0$ leads to the roots

Intermediate

A 17 and −18 B −17 and 18 C 18 and 17 D −18 and −17

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17 and −18

$(x+18)(x-17) = 0$.

Q86. Two consecutive positive integers have product 306. They are

Intermediate

A −18 and −17 B 16 and 19 C 17 and 18 D 15 and 20

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17 and 18

The roots are 17 and $-18$; the negative one is rejected because the integers are positive.

Q87. Why is one root rejected in a word problem?

Intermediate

A because it is negative B because the quantity it stands for cannot take that value C because it is a fraction D because the equation has only one root

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because the quantity it stands for cannot take that value

A length, a speed, an age or a count has its own restrictions; being negative is only the commonest reason.

Q88. Assertion (A): $x^2 + 6x + 9 = 0$ has two equal roots. Reason (R): It factorises as $(x + 3)^2 = 0$, and both factors give the same value of x.

Intermediate

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

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Both A and R are true, and R is the correct explanation of A

A perfect square factor is exactly what produces equal roots.

Q89. Assertion (A): Every quadratic equation can be solved by factorisation. Reason (R): Splitting the middle term needs two integers whose product is ac and whose sum is b, and such integers do not always exist.

Intermediate

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

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A is false but R is true

A is false and R explains why: $x^2 + x - 1 = 0$ has roots but no integer split.

Q90. Which equation has two equal roots?

Intermediate

A $x^2 - 6x + 9 = 0$ B $x^2 - 6x + 8 = 0$ C $x^2 - 9 = 0$ D $x^2 - 6x = 0$

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$x^2 - 6x + 9 = 0$

$x^2 - 6x + 9 = (x-3)^2$, a perfect square.

Q91. The roots of $3x^2 - 10x + 8 = 0$ are

Intermediate

A 2 and $\frac43$ B $-2$ and $-\frac43$ C 4 and $\frac23$ D $\frac12$ and 4

Show answer & explanation

2 and $\frac43$

$ac = 24$, $b = -10$, giving $-6$ and $-4$, so $(x-2)(3x-4) = 0$.

Q92. Two consecutive positive odd integers have product 143. The smaller is

Intermediate

A 13 B 9 C 11 D 7

Show answer & explanation

11

Odd integers differ by 2: $x^2 + 2x - 143 = 0$ gives 11 and $-13$; the negative root is rejected.

Q93. Which step should come FIRST when solving $6x^2 - 18x + 12 = 0$?

Intermediate

A find two numbers multiplying to 72 B set each bracket to zero C substitute a trial value D divide every term by 6

Show answer & explanation

divide every term by 6

Dividing gives $x^2 - 3x + 2 = 0$, whose split needs only a product of 2 rather than 72.

Q94. Solve by factorisation: $100x^2 - 20x + 1 = 0$.

Intermediate

Show answer & explanation

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Q95. Solve $2x^2 - 14x - 120 = 0$, dividing first.

Intermediate

Show answer & explanation

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Q96. Find two numbers whose sum is 27 and whose product is 182.

Intermediate

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Q97. Find two consecutive positive integers whose product is 306.

Intermediate

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Q98. Explain why $x = -32$ must be rejected when solving the train problem $x^2 - 8x - 1280 = 0$.

Intermediate

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Q99. (a) State the zero-product rule and explain why it allows a quadratic equation to be solved by factorisation. (b) Solve $2x^2 - 5x + 3 = 0$ by splitting the middle term, showing every step. (c) Verify both roots in the original equation.

Intermediate

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1½ marks — (a) the rule and the reasoning. 2½ marks — (b) the split, the grouping and both roots. 1 mark — (c) both verifications.

Q100. (a) Solve $x^2 + 4x + 4 = 0$ and say how many roots it has. (b) Solve $x^2 - 4 = 0$ and say how many roots it has. (c) Explain the difference between the two answers in terms of the factors.

Intermediate

Show answer & explanation

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2 marks — (a) the factorisation and 'two equal roots'. 1½ marks — (b) the factorisation and both roots. 1½ marks — (c) the explanation in terms of identical versus different factors.

Q101. For $2x^2 + x - 3 = 0$, the discriminant is

Intermediate

A 25 B −23 C 23 D −25

Show answer & explanation

25

$1 - 4(2)(-3) = 1 + 24 = 25$; the two negatives make the term positive.

Q102. For $3x^2 - 2x - 1 = 0$, the nature of the roots is

Intermediate

A two equal roots B no real roots C two distinct real roots D cannot be decided

Show answer & explanation

two distinct real roots

$D = 4 + 12 = 16 > 0$.

Q103. Using the formula on $3x^2 - 5x + 2 = 0$ gives the roots

Intermediate

A 1 and $\frac23$ B $-1$ and $-\frac23$ C $\frac53$ and $\frac23$ D 5 and 1

Show answer & explanation

1 and $\frac23$

$D = 1$, so $x = \frac{5 \pm 1}{6}$.

Q104. Applying the formula to $x^2 + x - 1 = 0$ gives

Intermediate

A $\frac{1 \pm \sqrt5}{2}$ B $\frac{-1 \pm \sqrt3}{2}$ C $\frac{-1 \pm \sqrt5}{2}$ D no real roots

Show answer & explanation

$\frac{-1 \pm \sqrt5}{2}$

$D = 1 + 4 = 5$, so $x = \frac{-1 \pm \sqrt5}{2}$.

Q105. The roots of a quadratic are rational exactly when D is

Intermediate

A any positive number B zero or a positive perfect square C negative D any even number

Show answer & explanation

zero or a positive perfect square

A non-square positive D leaves a surd in the answer.

Q106. Three equations share the terms $4x^2 - 12x$ but end in $+9$, $+8$ and $+10$. The one with a repeated root is

Intermediate

A $4x^2 - 12x + 8 = 0$ B $4x^2 - 12x + 9 = 0$ C $4x^2 - 12x + 10 = 0$ D none of the three

Show answer & explanation

$4x^2 - 12x + 9 = 0$

Their discriminants are 0, 16 and $-16$, and only a zero discriminant repeats a root.

Q107. For $2x^2 - 6x + 3 = 0$, the discriminant is

Intermediate

A −12 B 60 C 24 D 12

Show answer & explanation

12

$36 - 24 = 12$, positive but not a perfect square, so the roots are irrational.

Q108. The roots of $2x^2 - 6x + 3 = 0$ are

Intermediate

A $\frac{-3 \pm \sqrt3}{2}$ B $\frac{3 \pm \sqrt3}{2}$ C $\frac{6 \pm \sqrt{12}}{2}$ D $\frac{3 \pm \sqrt{12}}{2}$

Show answer & explanation

$\frac{3 \pm \sqrt3}{2}$

$x = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt3}{4} = \frac{3 \pm \sqrt3}{2}$.

Q109. Equal roots require the condition

Intermediate

A $b^2 > 4ac$ B $b^2 < 4ac$ C $b^2 = 4ac$ D $b = 2a$

Show answer & explanation

$b^2 = 4ac$

$D = 0$ rearranges to $b^2 = 4ac$.

Q110. Saying an equation with $D < 0$ has 'no solution' is

Intermediate

A correct B correct only if a is negative C correct only if c is negative D imprecise — it has no REAL roots

Show answer & explanation

imprecise — it has no REAL roots

The roots exist as complex numbers; the examinable phrase is 'no real roots'.

Q111. Assertion (A): If D is a positive perfect square, the equation can also be solved by factorisation. Reason (R): A perfect square discriminant makes both roots rational.

Intermediate

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

Show answer & explanation

Both A and R are true, and R is the correct explanation of A

Rational roots are exactly the ones an integer split can produce.

Q112. Assertion (A): For $x^2 - 4x + 4 = 0$ there is only one root. Reason (R): Its discriminant is zero.

Intermediate

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

Show answer & explanation

A is false but R is true

A is false — there are two EQUAL roots, both 2 — while R is true and is what produces that equality.

Q113. For $4x^2 + 3x - 1 = 0$, the discriminant is

Intermediate

A −7 B 7 C 25 D −25

Show answer & explanation

25

$9 - 4(4)(-1) = 9 + 16 = 25$.

Q114. Applying the formula to $4x^2 + 3x - 1 = 0$ gives

Intermediate

A $\frac14$ and $-1$ B $-\frac14$ and 1 C 4 and $-1$ D $\frac12$ and $-1$

Show answer & explanation

$\frac14$ and $-1$

$x = \frac{-3 \pm 5}{8}$.

Q115. An equation with discriminant 20 has roots that are

Intermediate

A real, distinct and rational B equal C not real D real, distinct and irrational

Show answer & explanation

real, distinct and irrational

20 is positive but not a perfect square.

Q116. Assertion (A): Knowing D tells you the roots themselves. Reason (R): D decides only how many real roots there are and whether they are rational.

Intermediate

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

Show answer & explanation

A is false but R is true

A is false — D gives the nature, not the values — and R states correctly what it does tell you.

Q117. Solve $3x^2 - 5x + 2 = 0$ using the quadratic formula.

Intermediate

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Q118. Solve $2x^2 - 6x + 3 = 0$ using the quadratic formula.

Intermediate

Show answer & explanation

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Q119. Solve $x^2 + x - 1 = 0$, and explain why factorisation would not have worked.

Intermediate

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Q120. Find the value of k for which $2x^2 + kx + 3 = 0$ has two equal roots.

Intermediate

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Q121. Can a rectangular park be laid out with an 80 m boundary enclosing 500 m²?

Intermediate

Show answer & explanation

—

Q122. Explain the difference between 'no real roots' and 'no solution'.

Intermediate

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Q123. (a) Define the discriminant and state what each of its three cases tells you. (b) Find the nature of the roots of $2x^2 - 4x + 3 = 0$ and of $3x^2 - 2x - 1 = 0$. (c) For the one that has real roots, find them using the quadratic formula.

Intermediate

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2 marks — (a) the definition and all three cases. 2 marks — (b) both discriminants and both conclusions. 1 mark — (c) both roots.

Q124. A train travels 480 km at a uniform speed. Had it been 8 km/h slower it would have taken 3 hours more. Taking the speed as x km/h, the equation is

Advanced

A $x^2 + 8x - 1280 = 0$ B $x^2 - 8x + 1280 = 0$ C $3x^2 - 8x - 480 = 0$ D $x^2 - 8x - 1280 = 0$

Show answer & explanation

$x^2 - 8x - 1280 = 0$

$\frac{480}{x-8} - \frac{480}{x} = 3$; clearing denominators gives $3840 = 3x^2 - 24x$.

Q125. In a right triangle the height is 7 cm short of the base, and the hypotenuse measures 13 cm. Taking the base as x cm, the equation is

Advanced

A $x^2 + 7x - 60 = 0$ B $x^2 - 7x + 60 = 0$ C $x^2 - 7x - 60 = 0$ D $2x^2 - 7x - 169 = 0$

Show answer & explanation

$x^2 - 7x - 60 = 0$

$x^2 + (x-7)^2 = 169$ gives $2x^2 - 14x - 120 = 0$; dividing by 2 gives $x^2 - 7x - 60 = 0$.

Q126. For $(x - 3)(2x + 1) = x(x + 5)$, the standard form is

Advanced

A $2x^2 - 10x - 3 = 0$ B $x^2 - 5x - 3 = 0$ C $3x^2 - 10x - 3 = 0$ D $x^2 - 10x - 3 = 0$

Show answer & explanation

$x^2 - 10x - 3 = 0$

The left expands to $2x^2 - 5x - 3$ and the right to $x^2 + 5x$, leaving $x^2 - 10x - 3 = 0$.

Q127. For which value of k is $kx^2 + 3x - 5 = 0$ NOT a quadratic equation?

Advanced

A 0 B 1 C 3 D −5

Show answer & explanation

0

The coefficient of $x^2$ must be non-zero, so only $k = 0$ fails.

Q128. If 2 is a root of $x^2 + kx - 6 = 0$, then k equals

Advanced

A −1 B 1 C 2 D 3

Show answer & explanation

1

Substituting gives $4 + 2k - 6 = 0$, so $2k = 2$.

Q129. A cottage industry makes x articles in a day. The cost of each in rupees is 3 more than twice the number made, and the day's total cost is ₹90. The equation is

Advanced

A $x^2 + 3x - 90 = 0$ B $2x^2 + 3x - 90 = 0$ C $2x^2 + 3x + 90 = 0$ D $2x^2 - 3x - 90 = 0$

Show answer & explanation

$2x^2 + 3x - 90 = 0$

Cost per article $2x + 3$, so $x(2x + 3) = 90$.

Q130. The sum of a positive number and its reciprocal is $\frac{26}{5}$. The equation is

Advanced

A $5x^2 + 26x + 5 = 0$ B $x^2 - 26x + 5 = 0$ C $5x^2 - 26x + 5 = 0$ D $5x^2 - 26x - 5 = 0$

Show answer & explanation

$5x^2 - 26x + 5 = 0$

$x + \frac1x = \frac{26}{5}$; multiplying by $5x$ gives $5x^2 + 5 = 26x$.

Q131. Assertion (A): $\frac{1}{x} + x = 4$ is a quadratic equation. Reason (R): Clearing the fraction leaves a term in $x^2$, a term in $x$ and a constant.

Advanced

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

Show answer & explanation

Both A and R are true, but R is not the correct explanation of A

Both are true, but simply naming the three terms does not decide the degree; it is the highest power being two that makes the equation quadratic.

Q132. Which situation does NOT lead to a quadratic equation?

Advanced

A a rectangle whose perimeter is 40 m and whose length is 3 m more than its breadth B a rectangle of area 528 m² whose length is one more than twice its breadth C two consecutive integers whose product is 306 D a number and its reciprocal adding to $\frac{26}{5}$

Show answer & explanation

a rectangle whose perimeter is 40 m and whose length is 3 m more than its breadth

A perimeter condition adds the sides rather than multiplying them, so it stays linear.

Q133. In a 'represent the situation' question, the final mark is usually for

Advanced

A simplifying the equation to standard form B drawing a diagram C solving the equation D checking the discriminant

Show answer & explanation

simplifying the equation to standard form

The question asks for the equation, and standard form is what completes it.

Q134. Rohan's mother is 26 years older than him. Three years from now the product of their ages will be 360. Represent this as a quadratic equation.

Advanced

Show answer & explanation

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Q135. A train covers 480 km at a uniform speed. Had the speed been 8 km/h less, the journey would have taken 3 hours longer. Represent this as a quadratic equation.

Advanced

Show answer & explanation

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Q136. A small pottery workshop turns out some number of pieces in a day. On one such day each piece cost, in rupees, 3 more than twice however many were made, and the day’s costs came to ₹90 altogether. (a) Taking the number of articles as x, write an expression for the cost of one article. (b) Form a quadratic equation for the situation. (c) Verify that $x = 6$ satisfies your equation, and state what it means.

Advanced

Show answer & explanation

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1 mark — (a) the cost expression. 2 marks — (b) the equation in standard form. 2 marks — (c) the substitution and its interpretation.

Q137. Represent each of these situations as a quadratic equation in standard form, defining your variable each time. (a) The sum of the squares of two consecutive positive integers is 365. (b) In a right triangle the height falls 7 cm short of the base, and the hypotenuse measures 13 cm. (c) The sum of a positive number and its reciprocal is $\frac{26}{5}$.

Advanced

Show answer & explanation

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1½ marks — (a) variable defined and equation simplified. 2 marks — (b) Pythagoras used and equation simplified. 1½ marks — (c) denominators cleared and equation in standard form.

Q138. The roots of $2x^2 + x - 528 = 0$ are

Advanced

A $\frac{33}{2}$ and $-16$ B 33 and $-16$ C $\frac{33}{2}$ and 16 D 16 and $-\frac{33}{2}$

Show answer & explanation

16 and $-\frac{33}{2}$

$ac = -1056$ and $b = 1$, giving 33 and $-32$; $(2x + 33)(x - 16) = 0$.

Q139. Solving $2x^2 + x - 528 = 0$ for the breadth of a plot gives a breadth of

Advanced

A 33 m B 16 m C 24 m D 22 m

Show answer & explanation

16 m

$2x^2 + x - 528 = 0$ gives $x = 16$ or $x = -\frac{33}{2}$; a breadth cannot be negative.

Q140. Factorised, $x^2 - 8x - 1280 = 0$ has roots

Advanced

A −40 and 32 B 32 and −40 C 40 and −32 D 64 and −20

Show answer & explanation

40 and −32

$ac = -1280$ and $b = -8$, giving $-40$ and 32, so $(x-40)(x+32) = 0$.

Q141. A train covers 480 km; at 8 km/h less it would take 3 hours more. Its speed is

Advanced

A 32 km/h B 40 km/h C 48 km/h D 60 km/h

Show answer & explanation

40 km/h

The equation $x^2 - 8x - 1280 = 0$ gives 40 and $-32$, and a speed cannot be negative.

Q142. Solving $x^2 - 7x - 60 = 0$ for the base of a right triangle gives a base of

Advanced

A 5 cm B 13 cm C 7 cm D 12 cm

Show answer & explanation

12 cm

$x^2 - 7x - 60 = 0$ gives 12 and $-5$; a length cannot be negative, so the base is 12 cm (and the altitude 5 cm).

Q143. Two numbers have sum 27 and product 182. They are

Advanced

A 13 and 14 B 12 and 15 C 11 and 16 D 10 and 17

Show answer & explanation

13 and 14

With the numbers as x and $27 - x$: $x(27 - x) = 182$ gives $x^2 - 27x + 182 = 0$, which factorises as $(x-13)(x-14) = 0$.

Q144. Solving $x^2 + x - 182 = 0$ for two consecutive positive integers, the smaller one is

Advanced

A 14 B 12 C 13 D 15

Show answer & explanation

13

$x^2 + x - 182 = 0$ gives 13 and $-14$; the integers are positive, so they are 13 and 14.

Q145. For which value of k does $x^2 - kx + 9 = 0$ factorise as a perfect square with positive k?

Advanced

A 3 B 9 C 6 D 18

Show answer & explanation

6

$(x - 3)^2 = x^2 - 6x + 9$, so $k = 6$.

Q146. A cottage industry made x articles in a day, each costing ₹$(2x + 3)$, for a total of ₹90. The number of articles was

Advanced

A 15 B 6 C 9 D 5

Show answer & explanation

6

$2x^2 + 3x - 90 = 0$ factorises as $(x - 6)(2x + 15) = 0$; a count cannot be negative, so $x = 6$.

Q147. Assertion (A): For the train problem, $-32$ is not an acceptable answer. Reason (R): Both roots of that equation satisfy it exactly.

Advanced

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

Show answer & explanation

Both A and R are true, but R is not the correct explanation of A

Both are true. That $-32$ satisfies the equation is precisely why it cannot be rejected on algebraic grounds - it is rejected because a speed cannot be negative.

Q148. Rehman's age x satisfies $x^2 - 4x - 21 = 0$. His age is

Advanced

A −3 years B 7 years C 21 years D 4 years

Show answer & explanation

7 years

$(x-7)(x+3) = 0$ gives 7 and $-3$; an age cannot be negative.

Q149. Solve by factorisation: $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$.

Advanced

Show answer & explanation

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Q150. In a right triangle the height falls 7 cm short of the base, and the hypotenuse measures 13 cm. Find the other two sides.

Advanced

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Q151. Why does $x^2 + x - 1 = 0$ resist factorisation?

Advanced

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Q152. Solve by factorisation: $3x^2 - 2\sqrt{6}x + 2 = 0$.

Advanced

Show answer & explanation

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Q153. A small pottery workshop turns out some number of pieces in a day. On one such day each piece cost, in rupees, 3 more than twice however many were made, and the total came to ₹90. (a) Form the quadratic equation. (b) Solve it by factorisation. (c) State the number of articles and the cost of each, explaining why you discard one root.

Advanced

Show answer & explanation

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1 mark — (a) the equation. 2 marks — (b) the split and both roots. 2 marks — (c) the rejection with its reason, and both final values.

Q154. A train covers 480 km without changing pace. Had it run 8 km/h slower, the same journey would have taken 3 hours longer. (a) Form the quadratic equation, taking the speed as x km/h. (b) Solve it by factorisation. (c) State the speed, explain the rejection, and find the time taken for the journey.

Advanced

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1½ marks — (a) the equation formed and cleared. 2 marks — (b) the factorisation and both roots. 1½ marks — (c) the rejection, the speed and the time.

Q155. For what value of k does $2x^2 + kx + 3 = 0$ have two equal roots?

Advanced

A $\pm 2\sqrt6$ B $2\sqrt6$ only C $\pm \sqrt{24}$ but not $-2\sqrt6$ D $\pm 6$

Show answer & explanation

$\pm 2\sqrt6$

$k^2 - 24 = 0$ gives $k^2 = 24$, so $k = \pm 2\sqrt6$; both signs are answers.

Q156. For what value of k does $kx^2 - 2kx + 6 = 0$ have equal roots?

Advanced

A 6 B 0 C 0 or 6 D 3

Show answer & explanation

6

$4k^2 - 24k = 0$ gives $k = 0$ or $k = 6$; $k = 0$ leaves $6 = 0$, which is not quadratic, so $k = 6$.

Q157. For what value of k does $x^2 - kx + 9 = 0$ have equal roots?

Advanced

A 6 only B $\pm 6$ C $\pm 3$ D 9

Show answer & explanation

$\pm 6$

$k^2 - 36 = 0$ gives $k = \pm 6$.

Q158. For which values of k does $x^2 + kx + 9 = 0$ have NO real roots?

Advanced

A $k > 6$ B $k < -6$ C $k = \pm 6$ D $-6 < k < 6$

Show answer & explanation

$-6 < k < 6$

$D = k^2 - 36 < 0$ requires $k^2 < 36$.

Q159. Can a rectangular park be laid out with an 80 m boundary enclosing 400 m²?

Advanced

A no, since $D < 0$ B yes, with sides 25 m and 16 m C yes — a 20 m square, since $D = 0$ D yes, with sides 40 m and 10 m

Show answer & explanation

yes — a 20 m square, since $D = 0$

$x^2 - 40x + 400 = 0$ has $D = 1600 - 1600 = 0$, giving the equal roots $x = 20$.

Q160. Can a rectangular park be laid out with an 80 m boundary enclosing 500 m²?

Advanced

A yes, 25 m by 20 m B no — the discriminant is $-400$ C yes, 50 m by 10 m D yes, but only as a square

Show answer & explanation

no — the discriminant is $-400$

$x^2 - 40x + 500 = 0$ has $D = 1600 - 2000 = -400 < 0$, so no such rectangle exists.

Q161. Can a mango grove be laid out as a rectangle twice as long as it is broad, covering 800 m²?

Advanced

A no, since $D < 0$ B yes, 80 m by 10 m C yes, 40 m by 20 m D yes, but only approximately

Show answer & explanation

yes, 40 m by 20 m

$2x^2 = 800$ gives $x^2 = 400$, so the breadth is 20 m and the length 40 m.

Q162. Two numbers add to 12 and multiply to 40. Such numbers

Advanced

A are 6 and 6 B are 10 and 2 C do not exist, since $D = -16$ D are 20 and −8

Show answer & explanation

do not exist, since $D = -16$

$x^2 - 12x + 40 = 0$ has $D = 144 - 160 = -16 < 0$, so no real pair works.

Q163. If one root of $x^2 + kx + 12 = 0$ is 3, the other root is

Advanced

A −4 B 4 C 9 D 12

Show answer & explanation

4

Substituting 3 gives $9 + 3k + 12 = 0$, so $k = -7$; then $x^2 - 7x + 12 = 0$ has roots 3 and 4.

Q164. Assertion (A): The equation $kx^2 + 4x + 1 = 0$ has equal roots when $k = 4$. Reason (R): Equal roots require $b^2 - 4ac = 0$, and here $16 - 4k = 0$.

Advanced

A Both A and R are true, and R is the correct explanation of A B Both A and R are true, but R is not the correct explanation of A C A is true but R is false D A is false but R is true

Show answer & explanation

Both A and R are true, and R is the correct explanation of A

$16 - 4k = 0$ gives $k = 4$, so R's condition produces exactly A's value.

Q165. A path's width x satisfies $x^2 + 5x - 36 = 0$. Its width is

Advanced

A 9 m B 4 m C −9 m D 36 m

Show answer & explanation

4 m

$D = 25 + 144 = 169$ gives $x = \frac{-5 \pm 13}{2}$, i.e. 4 or $-9$; a width cannot be negative.

Q166. Find the value of k for which $kx(x - 2) + 6 = 0$ has two equal roots.

Advanced

Show answer & explanation

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Q167. Can a rectangular park be laid out with an 80 m boundary enclosing 400 m²? If so, give its dimensions.

Advanced

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Q168. (a) State the quadratic formula and the condition under which it may be used. (b) Solve $2x^2 - 6x + 3 = 0$, leaving your answer in surd form. (c) Explain, using the discriminant, why this equation could not have been solved by splitting the middle term.

Advanced

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1 mark — (a) the formula and its condition. 2½ marks — (b) D, the substitution and the simplified surd. 1½ marks — (c) the reasoning from the non-square discriminant.

Q169. (a) Find the value of k for which $2x^2 + kx + 3 = 0$ has two equal roots. (b) Find the value of k for which $kx(x - 2) + 6 = 0$ has two equal roots. (c) Explain why one of the values you obtained in (b) must be rejected.

Advanced

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1½ marks — (a) $D = 0$ and both values of k. 2 marks — (b) standard form, D and both values. 1½ marks — (c) the rejection with the $a \ne 0$ reason.

Q170. A town planner is asked to lay out rectangular parks, each with a perimeter of 80 m. (a) Taking the length as x metres, write an expression for the breadth and hence for the area. (b) Determine whether a park of area 400 m² is possible, and give its dimensions if it is. (c) Determine whether a park of area 500 m² is possible, justifying your answer with the discriminant.

Advanced

Show answer & explanation

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1 mark — (a) breadth and area. 2 marks — (b) the equation, $D = 0$ and the dimensions. 2 marks — (c) the equation, the negative discriminant and the conclusion.

Quadratic Equations · Class 10 Mathematics: 170 MCQs with answers

Warm-up · 54 questions

Building up · 69 questions

Topper level · 47 questions

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