Quadratic Equations

A · $x^2 + 3x + 2 = 0$

A quadratic equation is of the form $ax^2 + bx + c = 0$ with $a \neq 0$. Only $x^2 + 3x + 2 = 0$ fits this form.

A · $ax^2 + bx + c = 0$

The standard form is $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$.

B · $b^2 - 4ac$

The discriminant is $D = b^2 - 4ac$, which determines the nature of roots.

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

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

B · $x = 2$ or $x = -2$

$x^2 - 4 = 0$ gives $x^2 = 4$, so $x = \pm 2$.

C · $x = 2$ or $x = -2$

$2x^2 - 8 = 0$ gives $x^2 = 4$, so $x = \pm 2$.

B · False

The discriminant is $2^2 - 4 \cdot 1 \cdot 1 = 0$, so the roots are equal, not distinct.

A · True

The quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ works for all quadratic equations.

C · $x = 2$ or $x = 3$

Factorizing: $(x-2)(x-3)=0$, so $x=2$ or $x=3$.

D · Real and equal

When $D = 0$, the quadratic equation has two equal real roots.

D · $-\frac{b}{a}$

The sum of roots is $-\frac{b}{a}$.

A · $\frac{c}{a}$

The product of roots is $\frac{c}{a}$.

D · Not real

When $D < 0$, the quadratic equation has no real roots.

C · 2

Factorizing: $(x-2)(x-5)=0$, so roots are 2 and 5. Hence 2 is a root.

C · Real and equal

Discriminant $D = 6^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0$, so roots are real and equal.