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

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Question

(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

Vedclass · Accepted Answer

-$2$

Vedclass · Answer

(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,$a\alpha = -\alpha^3 - 1$.Substituting this into the second equation: $\alpha^4 + \alpha(a\alpha^2) + 1 = 0$ is not direct,so multiply the first by $\alpha$: $\alpha^4 + a\alpha^2 + \alpha = 0$.Subtracting this from the second equation: $(\alpha^4 + a\alpha^2 + 1) - (\alpha^4 + a\alpha^2 + \alpha) = 0$.This gives $1 - \alpha = 0$,so $\alpha = 1$.Substituting $\alpha = 1$ into the first equation: $1^3 + a(1) + 1 = 0$.$1 + a + 1 = 0$,which implies $a = -2$.

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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