The equations x^2-ax+b=0 and x^2+bx-a=0 have | vedclass

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(C) Let $\alpha$ be the common root of the equations $x^2-ax+b=0$ and $x^2+bx-a=0$.Then,$\alpha^2-a\alpha+b=0$ and $\alpha^2+b\alpha-a=0$.Subtracting

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$a+b=0$ or $a-b=1$

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(C) Let $\alpha$ be the common root of the equations $x^2-ax+b=0$ and $x^2+bx-a=0$.Then,$\alpha^2-a\alpha+b=0$ and $\alpha^2+b\alpha-a=0$.Subtracting the two equations:$(\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 implies either $a+b=0$ or $\alpha=1$.If $\alpha=1$ is a root,then $1^2-a(1)+b=0$,which gives $1-a+b=0$,or $a-b=1$.Thus,the condition for a common root is $a+b=0$ or $a-b=1$.

The equations $x^2-ax+b=0$ and $x^2+bx-a=0$ have a common root. Then:

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