Conditions on polynomials with common roots.

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If one root of the equation $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$ is common and $a \ne b$ then:

The options are as follows: $$\begin{array}{ll} (A)\quad& a + b = 0\\ (B)& a + b = -1\\ (C)& a - b = 1\\ (D)& a + b = 1 \end{array}$$

Idk how to solve this, please help me.

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Let $c$ be the common root. Then $$c^2+ac+b=0$$ $$c^2+bc+a=0$$ so $ac+b=bc+a$ or $(a-b)c=a-b$. Since $a\neq b$, $c$ must be $1$.

Now we have $a+b=-1$.

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Hint: Substitute the common root and subtract.

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$x^2+ax+b=x^2+bx+a$ => $(a-b)(x-1)=0$. Since $a \neq b$, $x=1$. Thus $1^2+a+b=0$. Hence, $a+b=-1$.

Thus, C is the answer.