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If $\alpha$ and $\beta$ are roots of the equation, $x^{2}-4\sqrt{2}kx+2e^{4ln\: k}-1=0$ for some k

and $\alpha ^{2}+\beta ^{2}=66$ then $\alpha ^{3}+\beta ^{3}$ is equal to:

Option 1)
$248\sqrt{2}$

Option 2)
$280\sqrt{2}$

Option 3)
$-32\sqrt{2}$

Option 4)
$-280\sqrt{2}$

Answers (2)

best_answer

As we have learned

Sum of Roots in Quadratic Equation -

$\alpha +\beta = \frac{-b}{a}$

- wherein

$\alpha \: and\beta$ are root of quadratic equation

$ax^{2}+bx+c=0$

$a,b,c\in C$

Product of Roots in Quadratic Equation -

$\alpha \beta = \frac{c}{a}$

- wherein

$\alpha \: and\ \beta$ are roots of quadratic equation:

$ax^{2}+bx+c=0$

$a,b,c\in C$

$x^2-4\sqrt2kx+2e^{lnk^4}-1= 0$

$\Rightarrow x^2-4\sqrt2kx+2k^4-1=0$

$\alpha +\beta = 4\sqrt2 k$ and

$\alpha\beta =2k^4-1$

Now , $\alpha ^2+\beta ^2= (\alpha +\beta )^{2}-2\alpha \beta$

$\Rightarrow 66= 32k^2-4 k^4+2$

$\Rightarrow 4k^4-32k^2+64 = 0$

$\Rightarrow k^4-8k^2+16 = 0$

$\Rightarrow (k^2-4)^= 0$

$\Rightarrow k= \pm 2$

k= 2 acceptable for ln k to be definite

$\therefore \alpha ^2+\beta ^2= (\alpha +\beta )^3-3\alpha \beta (\alpha +\beta )= (4\sqrt2 \times 2)^3-3(32-1)(4\sqrt2\times 2)$ $= 1024 \sqrt2- 744\sqrt2= 280 \sqrt 2$

Option 1)

$248\sqrt{2}$

Option 2)

$280\sqrt{2}$

Option 3)

$-32\sqrt{2}$

Option 4)

$-280\sqrt{2}$

Posted by

Himanshu

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