If alpha is the common root of the quadratic | vedclass

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Question

(A) Since $\alpha$ is the common root,it satisfies both equations:
$1) \alpha^2 - 5\alpha + 4a = 0 \implies 4a = 5\alpha - \alpha^2$
$2) \alpha^2 - 2a

Vedclass · Accepted Answer

$260$

Vedclass · Answer

(A) Since $\alpha$ is the common root,it satisfies both equations:
$1) \alpha^2 - 5\alpha + 4a = 0 \implies 4a = 5\alpha - \alpha^2$
$2) \alpha^2 - 2a\alpha - 8 = 0$
Substitute $2a = \frac{5\alpha - \alpha^2}{2}$ into the second equation:
$\alpha^2 - (\frac{5\alpha - \alpha^2}{2})\alpha - 8 = 0$
$2\alpha^2 - 5\alpha^2 + \alpha^3 - 16 = 0$
$\alpha^3 - 3\alpha^2 - 16 = 0$
By testing integer roots,$\alpha = 4$ satisfies the equation: $64 - 3(16) - 16 = 64 - 48 - 16 = 0$.
For $\alpha = 4$,$4a = 5(4) - 16 = 4 \implies a = 1$.
The expression is $\alpha^4 - \alpha^3 + 68 = 4^4 - 4^3 + 68 = 256 - 64 + 68 = 260$.

If $\alpha$ is the common root of the quadratic equations $x^2-5x+4a=0$ and $x^2-2ax-8=0$,where $a \in R$,then the value of $\alpha^4-\alpha^3+68$ is

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