Question

If the equation $x^2 - ax + b = 0$ and $x^2 + bx - a = 0$ have a common root, then-

• A. a = b
• B. a + b = 0
• C. a - b = 1
• D. a - b + 1 = 0

Answer 1

Answer: (B)

a + b = 0

Answer 2

Let the common root of the two quadratic equations $x^2 - ax + b = 0$ and $x^2 + bx - a = 0$ be $\alpha$.

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

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

Subtract equation (2) from equation (1):

$(\alpha^2 - a\alpha + b) - (\alpha^2 + b\alpha - a) = 0$ $\alpha^2 - a\alpha + b - \alpha^2 - b\alpha + a = 0$ $-a\alpha - b\alpha + b + a = 0$ $-(a+b)\alpha + (a+b) = 0$ $(a+b)(1 - \alpha) = 0$

This equation implies two possibilities:

Case 1: $a+b = 0$

If $a+b=0$, then $b=-a$. Substituting this into the original equations:

Equation 1: $x^2 - ax - a = 0$ Equation 2: $x^2 - ax - a = 0$

In this case, the two equations are identical, meaning they have both roots in common, and thus certainly have "a common root". So, $a+b=0$ is a valid condition.

Case 2: $1 - \alpha = 0 \implies \alpha = 1$

If $\alpha = 1$ is the common root, substitute $\alpha = 1$ into equation (1):

$1^2 - a(1) + b = 0$ $1 - a + b = 0$ $a - b = 1$

So, if the common root is $1$, then $a-b=1$ is a valid condition.

Therefore, if the two equations have a common root, then either $a+b=0$ or $a-b=1$. Both (B) and (C) are presented as options. If it's a single-choice question, it implies one of these must be chosen. Both are mathematically correct derivations. Option (B) implies that the equations are identical, thus having common roots.