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If , are real and ^2, -^2 are the roots of a^2 x^2 + x+1-a^2=

Question

If $α, β$ are real and $α^{2}, - β^{2}$ are the roots of $a^{2} x^{2} + x + 1 - a^{2} = 0; (a > 1)$ , then $β^{2} =$

a^{2}

1

1 - a^{2}

1 + a^{2}

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Liam Discussed

α, β

are real and

α^{2}, - β^{2}

are the roots of

a^{2} x^{2} + x + 1 - a^{2} = 0; (a > 1)

, then

β^{2} =

9 mins ago

Discuss this question LIVE

9 mins ago

Text solutionVerified

α^{2}, - β^{2}

are the roots of

a^{2} x^{2} + x + 1 - a^{2} = 0

\Rightarrow α^{2} - β^{2} = - \frac{1}{a ^{2}},

And

\Rightarrow α^{2} (- β^{2}) = \frac{1 - a ^{2}}{a ^{2}}

(α^{2} + β^{2})^{2} = (α^{2} - β^{2})^{2} + 4 α^{2} β^{2}

= \frac{1}{a ^{4}} + 4 (\frac{a ^{2} - 1}{a ^{2}}) = \frac{1}{a ^{4}} + 4 - \frac{4}{a ^{2}}

= (2 - \frac{1}{a ^{2}})^{2} \Rightarrow α^{2} + β^{2} = 2 - \frac{1}{a ^{2}}

β^{2} = \frac{1}{2} [(α^{2} + β^{2}) - (α^{2} - β^{2})]

= \frac{1}{2} [2 - \frac{1}{a ^{2}} + \frac{1}{a ^{2}}] = 1

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Question Text	If $α, β$ are real and $α^{2}, - β^{2}$ are the roots of $a^{2} x^{2} + x + 1 - a^{2} = 0; (a > 1)$ , then $β^{2} =$
Answer Type	Text solution:1
Upvotes	150