The equations x^{2}+x+a=0 and x^{2}+ax+1=0 ha | vedclass

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(B) Let $\alpha$ be the common root.Then,$\alpha^{2}+\alpha+a=0$ ... $(i)$And $\alpha^{2}+a\alpha+1=0$ ... (ii)Subtracting (ii) from $(i)$,we get:$(\a

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for exactly one value of $a$

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(B) Let $\alpha$ be the common root.Then,$\alpha^{2}+\alpha+a=0$ ... $(i)$And $\alpha^{2}+a\alpha+1=0$ ... (ii)Subtracting (ii) from $(i)$,we get:$(\alpha^{2}+\alpha+a) - (\alpha^{2}+a\alpha+1) = 0$$\alpha(1-a) + (a-1) = 0$$\alpha(1-a) - (1-a) = 0$$(1-a)(\alpha-1) = 0$This implies $a=1$ or $\alpha=1$.Case $1$: If $a=1$,the equations become $x^{2}+x+1=0$,which has no real roots.Case $2$: If $\alpha=1$,substituting into $(i)$ gives $1^{2}+1+a=0$,so $a=-2$.For $a=-2$,the equations are $x^{2}+x-2=0$ and $x^{2}-2x+1=0$.The roots of $x^{2}+x-2=0$ are $x=1, -2$.The roots of $x^{2}-2x+1=0$ are $x=1, 1$.The common root is $x=1$,which is real.Thus,there is exactly one value of $a$ $(a=-2)$ for which the equations have a common real root.

The equations $x^{2}+x+a=0$ and $x^{2}+ax+1=0$ have a common real root.

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