Question

Let $a, b$ and $c$ are real numbers such that $4a + 2b + c = 0$ and $ab > 0$. Then the equation $ax^2 + bx + c = 0$ has:

• A. Unequal real roots with same sign
• B. Imaginary roots
• C. Equal roots
• D. Real roots with opposite signs

Answer 1

Answer: (D)

Real roots with opposite signs

Answer 2

The condition $4a + 2b + c = 0$ implies that $x=2$ is a root of the equation $ax^2 + bx + c = 0$, because $a(2)^2 + b(2) + c = 4a + 2b + c = 0$.

Let the roots be $x_1$ and $x_2$. We know $x_1 = 2$. From Vieta's formulas: Sum of roots: $x_1 + x_2 = 2 + x_2 = -\frac{b}{a}$ Product of roots: $x_1 \cdot x_2 = 2 \cdot x_2 = \frac{c}{a}$

Given $ab > 0$, this means $a$ and $b$ have the same sign, so $\frac{b}{a} > 0$. From the sum of roots: $2 + x_2 = -\frac{b}{a}$. Since $\frac{b}{a} > 0$, then $-\frac{b}{a} < 0$. Therefore, $2 + x_2 < 0$, which implies $x_2 < -2$.

Since one root is $2$ (positive) and the other root $x_2$ is less than $-2$ (negative), the roots have opposite signs. Both roots are real and unequal.