If one of the roots of the equations x^2 + ax | vedclass

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(B) 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\alp

Vedclass · Accepted Answer

$-1$

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(B) 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: $(\alpha^2 + a\alpha + b) - (\alpha^2 + b\alpha + a) = 0$.$\alpha(a - b) - (a - b) = 0$.$(a - b)(\alpha - 1) = 0$.Since the roots are coincident,we assume $a 
eq b$,so $\alpha - 1 = 0$,which gives $\alpha = 1$.Substituting $\alpha = 1$ into the first equation: $1^2 + a(1) + b = 0$.$1 + a + b = 0$.Therefore,$a + b = -1$.

If one of the roots of the equations $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$ is coincident,then the numerical value of $(a + b)$ is

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