The value of a for which the equations x^3+ax | vedclass

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(A) Let the common root be $\alpha$. Then $\alpha^3+a\alpha+1=0$ and $\alpha^4+a\alpha^2+1=0$.From the first equation,$a\alpha = -\alpha^3-1$,so $a =

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-$2$

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(A) Let the common root be $\alpha$. Then $\alpha^3+a\alpha+1=0$ and $\alpha^4+a\alpha^2+1=0$.From the first equation,$a\alpha = -\alpha^3-1$,so $a = -\alpha^2 - \frac{1}{\alpha}$.Substitute $a$ into the second equation: $\alpha^4 + (-\alpha^2 - \frac{1}{\alpha})\alpha^2 + 1 = 0$.$\alpha^4 - \alpha^4 - \alpha + 1 = 0$.This gives $\alpha = 1$.Substituting $\alpha = 1$ into $x^3+ax+1=0$,we get $1^3+a(1)+1=0$,which implies $1+a+1=0$,so $a = -2$.

The value of $a$ for which the equations $x^3+ax+1=0$ and $x^4+ax^2+1=0$ have a common root is

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