Let be the roots of the equation x 2 + 2 2 x - 1 = 0 . The quadratic equation, whose roots are , is: [2024]

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Solution

Q.
Let $α, β$ be the roots of the equation $x^{2} + 2 \sqrt{2} x - 1 = 0 .$ The quadratic equation, whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6}),$ is: [2024]

1 $x^{2} - 195 x + 9466 = 0$

2 $x^{2} - 180 x + 9506 = 0$

3 $x^{2} - 195 x + 9506 = 0$

4 $x^{2} - 190 x + 9466 = 0$

Ans.
(3)

We have, $x^{2} + 2 \sqrt{2 x} - 1 = 0$

$∴$ $α + β = - 2 \sqrt{2}$ and $α β = - 1$

Now, $α^{2} + β^{2} = {(α + β)}^{2} - 2 α β = {(- 2 \sqrt{2})}^{2} - 2 (- 1)$

$= 8 + 2 = 10$

Also, $α^{4} + β^{4} = {(α^{2} + β^{2})}^{2} - 2 {(α β)}^{2} = (10)$ ²-2(1)

$= 100 - 2 = 98$

and $α^{6} + β^{6} = {(α^{2} + β^{2})}^{3} - 3 α^{2} β^{2} (α^{2} + β^{2}) = {(10)}^{3} - 3 (1) (10)$

$= 1000 - 30 = 970$

So, $\frac{1}{10} (α^{6} + β^{6}) = 97$

Hence, equation whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6})$ is $x^{2} - (98 + 97) x + 98 \times 97 = 0$ i.e., $x^{2} - 195 x + 9506 = 0$

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