If-a-b-R-satisfy-a-4-b-4-6a-2-b-2-9-and-ab-a-b-a-b-11-then-a-2-b-2-

Question Number 151198 by liberty last updated on 19/Aug/21

If a, b \in R satisfy a^{4} + b^{4} - 6 a^{2} b^{2} = 9 a n d

a b (a - b) (a + b) = - 11 then a^{2} + b^{2} = ?

Answered by EDWIN88 last updated on 19/Aug/21

(1) a^{4} + b^{4} - 6 a^{2} b^{2} = 9

{(a^{2} - b^{2})}^{2} - 4 a^{2} b^{2} = 9

\underset{―}{{(a^{2} - b^{2})}^{2} = 4 {(a b)}^{2} + 9}

(2) a b (a - b) (a + b) = - 11

a b (a^{2} - b^{2}) = - 11

\underset{―}{a^{2} - b^{2} = - \frac{11}{a b}}

(1) = (2)

\Rightarrow \frac{121}{{(a b)}^{2}} = 4 {(a b)}^{2} + 9; l e t {(a b)}^{2} = X

\Rightarrow 4 X^{2} + 9 X - 121 = 0

\Rightarrow X = \frac{- 9 + \sqrt{2017}}{8}

i n s e r t i n g t o e q u a t i o n (1)

w e g e t a^{4} + b^{4} - 6 a^{2} b^{2} = 9

\Rightarrow {(a^{2} + b^{2})}^{2} - 8 {(a b)}^{2} = 9

\Rightarrow a^{2} + b^{2} = \sqrt{8 (\frac{\sqrt{2017} - 9}{8}) + 9}

\Rightarrow a^{2} + b^{2} = \sqrt{\sqrt{2017}} = \sqrt[4]{2017} .

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