If x^2-3ax+14=0 and x^2+2ax-16=0 have a commo | vedclass

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(B) Let the common root be $\alpha$. Then,$\alpha^2-3a\alpha+14=0$ and $\alpha^2+2a\alpha-16=0$. Subtracting the two equations: $(\alpha^2-3a\alpha+14

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) Let the common root be $\alpha$. Then,$\alpha^2-3a\alpha+14=0$ and $\alpha^2+2a\alpha-16=0$. Subtracting the two equations: $(\alpha^2-3a\alpha+14) - (\alpha^2+2a\alpha-16) = 0$ $-5a\alpha + 30 = 0 $ $\Rightarrow 5a\alpha = 30 $ $\Rightarrow \alpha = \frac{6}{a}$. Substituting $\alpha = \frac{6}{a}$ into the first equation: $(\frac{6}{a})^2 - 3a(\frac{6}{a}) + 14 = 0$ $\frac{36}{a^2} - 18 + 14 = 0$ $\frac{36}{a^2} = 4 \Rightarrow a^2 = 9$. Now,$a^4+a^2 = (a^2)^2 + a^2 = (9)^2 + 9 = 81 + 9 = 90$.

If $x^2-3ax+14=0$ and $x^2+2ax-16=0$ have a common root,then $a^4+a^2=$

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