If x² + 2ax + b = 0 and x² + 2bx + a = 0 have a common root and a ≠ b, prove that a + b = −1/4.

Question

ScanSolve AI Tutor · Accepted Answer

Here's the proof for the given statement.

Let $\alpha$ be the common root of the two quadratic equations.
Since $\alpha$ is a common root, it must satisfy both equations:
$\alpha^2 + 2a\alpha + b = 0 \quad (1)$ $\alpha^2 + 2b\alpha + a = 0 \quad (2)$
Subtract equation (2) from equation (1).
$(\alpha^2 + 2a\alpha + b) - (\alpha^2 + 2b\alpha + a) = 0$ $2a\alpha + b - 2b\alpha - a = 0$ $2\alpha(a - b) - (a - b) = 0$
Factor out $(a - b)$ :
$(a - b)(2\alpha - 1) = 0$
Solve for $\alpha$ using the given condition.
We are given that $a \neq b$ , which means $a - b \neq 0$ .
For the product $(a - b)(2\alpha - 1)$ to be zero, the second factor must be zero:
$2\alpha - 1 = 0$ $2\alpha = 1$ $\alpha = \frac{1}{2}$
Substitute the value of $\alpha$ back into one of the original equations.
Substitute $\alpha = \frac{1}{2}$ into equation (1):
$\left(\frac{1}{2}\right)^2 + 2a\left(\frac{1}{2}\right) + b = 0$ $\frac{1}{4} + a + b = 0$
Rearrange the equation to prove the required statement.
$a + b = -\frac{1}{4}$
This proves the statement.

If x² + 2ax + b = 0 and x² + 2bx + a = 0 have a common root and a ≠ b, prove that a + b = -1/4.