Question

If $x^2 + ax + 10 = 0$ and $x^2 + bx - 10 = 0$, have a common root, then $a^2 - b^2$ is equal to

• A. 10
• B. 20
• C. 30
• D. 40

Answer 1

Answer: (D)

40

Answer 2

Let the common root of the two quadratic equations be $\alpha$.

Since $\alpha$ is a common root, it must satisfy both equations:

1. $\alpha^2 + a\alpha + 10 = 0$
2. $\alpha^2 + b\alpha - 10 = 0$

Subtracting equation (2) from equation (1): $(\alpha^2 + a\alpha + 10) - (\alpha^2 + b\alpha - 10) = 0$

$\alpha^2 + a\alpha + 10 - \alpha^2 - b\alpha + 10 = 0$

$(a - b)\alpha + 20 = 0$

$(a - b)\alpha = -20 \quad \ldots (3)$

Adding equation (1) and equation (2): $(\alpha^2 + a\alpha + 10) + (\alpha^2 + b\alpha - 10) = 0$

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

$2\alpha^2 + (a + b)\alpha = 0$

$\alpha(2\alpha + a + b) = 0$

This implies two possibilities:

Case 1: $\alpha = 0$

If $\alpha = 0$, substitute it into equation (1):

$0^2 + a(0) + 10 = 0$

$10 = 0$

This is a contradiction, so $\alpha$ cannot be $0$.

Case 2: $2\alpha + a + b = 0$

From this, we get:

$a + b = -2\alpha \quad \ldots (4)$

We need to find the value of $a^2 - b^2$. We know that $a^2 - b^2 = (a - b)(a + b)$.

From equation (3), since $\alpha \neq 0$:

$a - b = -\frac{20}{\alpha}$

Substitute the expressions for $(a - b)$ and $(a + b)$ into the identity for $a^2 - b^2$:

$a^2 - b^2 = \left(-\frac{20}{\alpha}\right) (-2\alpha)$

$a^2 - b^2 = \frac{(-20)(-2)\alpha}{\alpha}$

$a^2 - b^2 = \frac{40\alpha}{\alpha}$

$a^2 - b^2 = 40$