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Quadratics Core: 17 MCQs with answers

All 17 multiple-choice questions from this kit, free and with no sign-up.

Full kit · everything Flashcards · 17 Quick Revision · 19

Warm-up · 7 questions

Q1. What is the standard form of a quadratic equation?

Beginner
A $ax^3 + bx^2 + cx + d = 0$ B $a^2x + b = 0$ C $ax + b = 0$ D $ax^2 + bx + c = 0$
Show answer & explanation

$ax^2 + bx + c = 0$

A quadratic equation is a polynomial equation of degree 2, so its standard form is $ax^2 + bx + c = 0$ where $a \neq 0$.

Q2. In the quadratic equation $3x^2 - 5x + 2 = 0$, what is the value of $a$?

Beginner
A 2 B 0 C -5 D 3
Show answer & explanation

3

In the standard form $ax^2 + bx + c = 0$, $a$ is the coefficient of $x^2$, which is 3 here.

Q3. Which of the following is a quadratic equation?

Beginner
A $x + \frac{1}{x} = 0$ B $2x + 5 = 0$ C $x^2 + 3x - 1 = 0$ D $x^3 - 2 = 0$
Show answer & explanation

$x^2 + 3x - 1 = 0$

A quadratic equation must have degree 2. Only $x^2 + 3x - 1 = 0$ fits that form.

Q4. Solve: $x^2 - 9 = 0$

Beginner
A $x = 0$ or $x = 9$ B $x = 9$ only C $x = 3$ only D $x = 3$ or $x = -3$
Show answer & explanation

$x = 3$ or $x = -3$

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

Q5. The roots of $x^2 - 5x + 6 = 0$ are:

Beginner
A -2 and -3 B 1 and 6 C 2 and 3 D -1 and -6
Show answer & explanation

2 and 3

Factoring gives $(x-2)(x-3)=0$, so roots are 2 and 3.

Q6. The equation $x^2 + 2x + 1 = 0$ has two distinct real roots.

Beginner
A True B False
Show answer & explanation

False

The discriminant is $4 - 4 = 0$, so it has exactly one real root (a repeated root).

Q7. In a quadratic equation, the coefficient $a$ can be zero.

Beginner
A True B False
Show answer & explanation

False

If $a=0$, the equation becomes linear, not quadratic. The definition requires $a \neq 0$.

Building up · 9 questions

Q8. What is the discriminant of the quadratic equation $x^2 - 4x + 4 = 0$?

Intermediate
A 0 B 8 C -16 D 16
Show answer & explanation

0

The discriminant is $b^2 - 4ac$. Here $a=1$, $b=-4$, $c=4$, so $(-4)^2 - 4(1)(4) = 16 - 16 = 0$.

Q9. How many real roots does a quadratic equation have if its discriminant is negative?

Intermediate
A 2 B Infinite C 0 D 1
Show answer & explanation

0

A negative discriminant means the equation has no real roots; the roots are complex conjugates.

Q10. What is the sum of the roots of the equation $2x^2 - 6x + 3 = 0$?

Intermediate
A 6 B 1.5 C 3 D -3
Show answer & explanation

3

For $ax^2 + bx + c = 0$, sum of roots = $-b/a$. Here $-(-6)/2 = 3$.

Q11. What is the product of the roots of the equation $x^2 + 5x - 14 = 0$?

Intermediate
A -5 B -14 C 14 D 5
Show answer & explanation

-14

Product of roots = $c/a$. Here $c = -14$ and $a=1$, so product is $-14$.

Q12. Which method is most efficient for solving $x^2 + 6x + 9 = 0$?

Intermediate
A Graphing B Completing the square C Factoring (perfect square) D Quadratic formula
Show answer & explanation

Factoring (perfect square)

The equation is $(x+3)^2 = 0$, so factoring is straightforward and efficient.

Q13. What is the value of the discriminant for $x^2 + 2x + 5 = 0$?

Intermediate
A 16 B -16 C 0 D 24
Show answer & explanation

-16

Discriminant $= 2^2 - 4(1)(5) = 4 - 20 = -16$.

Q14. Which of the following equations has exactly one real root?

Intermediate
A $x^2 - 2x + 1 = 0$ B $x^2 - 4 = 0$ C $x^2 - 3x + 2 = 0$ D $x^2 + 1 = 0$
Show answer & explanation

$x^2 - 2x + 1 = 0$

A quadratic has one real root when discriminant is zero. For $x^2 - 2x + 1$, discriminant $= 4 - 4 = 0$.

Q15. If one root of $x^2 - 5x + k = 0$ is 2, what is the value of $k$?

Intermediate
A -10 B -6 C 6 D 10
Show answer & explanation

6

Substitute $x=2$: $4 - 10 + k = 0$, so $k = 6$.

Q16. The equation $x^2 + 1 = 0$ has:

Intermediate
A Infinite roots B No real roots C Two real roots D One real root
Show answer & explanation

No real roots

Discriminant $= 0 - 4(1)(1) = -4 < 0$, so no real roots.

Topper level · 1 questions

Q17. Solve using the quadratic formula: $x^2 + 4x + 1 = 0$

Advanced
A $x = -2 \pm \sqrt{3}$ B $x = -4 \pm \sqrt{3}$ C $x = 2 \pm \sqrt{3}$ D $x = -2 \pm \sqrt{5}$
Show answer & explanation

$x = -2 \pm \sqrt{3}$

Using $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1, b=4, c=1$ gives $x = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}$.

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