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Algebra of Limits

If α ,β are...

Question

If $α, β$ are roots of the quadratic equation $x^{2} - x - 1 = 0,$ then the quadratic equation whose roots are $\frac{1 + α}{2 - α}, \frac{1 + β}{2 - β}$ is

z^{2} + z + 1 = 0

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z^{2} - 7 z + 1 = 0

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z^{2} + 7 z + 1 = 0

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z^{2} + 7 z - 1 = 0

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Solution

The correct option is B $z^{2} - 7 z + 1 = 0$
$x^{2} - x - 1 = 0$
$α + β = 1$
and $α . β = - 1$
Now,
$\frac{1 + α}{2 - α} + \frac{1 + β}{2 - β}$
$= \frac{2 - β + 2 α - α β + 2 - α + 2 β - α β}{4 - 2 (α + β) + α . β}$
$= \frac{4 + α + β - 2 α . β}{4 - 2 (α + β) + α . β}$
$= \frac{4 + 1 + 2}{4 - 2 - 1}$
$= 7$
And
$\frac{1 + α}{2 - α} \frac{1 + β}{2 - β}$
$= \frac{1 + (α + β) + α β}{4 - 2 (α + β) + α . β}$
$= \frac{1 + 1 - 1}{4 - 2 - 1}$
$= 1$
Hence, the equation is
$x^{2} - 7 x + 1 = 0$ .

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If α,β are roots of the quadratic equation x2−x−1=0, then the quadratic equation whose roots are 1+α2−α,1+β2−β is

If $α, β$ are roots of the quadratic equation $x^{2} - x - 1 = 0,$ then the quadratic equation whose roots are $\frac{1 + α}{2 - α}, \frac{1 + β}{2 - β}$ is