Ashifur August 8, 2023 - 09:18
Problem: Find the value of x$^{4}$+$\frac{1}{x^{4}}$,if x=$\sqrt{5-\sqrt{4}}$

View all: JANATA BANK ( EO-FA) | WRITTEN QUESTION (MATH) SOLVE | 2015

Correct Answer: The required value, x$^{4}$+ $\frac{1}{x^{4}}$= 322.

Explanation:

Given that, x = $\sqrt{5}-\sqrt{4}$……………………….. (i)

Now, Reversing Eqution …………………………..(i)

$\frac{1}{x}$= $\frac{1}{\sqrt{5-\sqrt{4}}}$

=> $\frac{1}{x}$ = $\frac{1(\sqrt{5+\sqrt{4}})}{(\sqrt{5-\sqrt{4}})(\sqrt{5+\sqrt{4}})}$

=> $\frac{1}{x}$ = $\frac{\sqrt{5+\sqrt{4}}}{(\sqrt{5})^{2}-(\sqrt{4})^{2}}$

=> $\frac{1}{x}$ = $\frac{\sqrt{5+\sqrt{4}}}{5-4}$

=> $\frac{1}{x}$ = $\sqrt{5}+\sqrt{4}$…………………..(ii)

Now, {(i) + (17)} we have,

$x+\frac{1}{x}=(\sqrt{5}-\sqrt{4})+(\sqrt{5}+\sqrt{4})$

=> $x+\frac{1}{x} = \sqrt{5}-\sqrt{4}+\sqrt{5}+\sqrt{4}$

=> $x+\frac{1}{x} = 2\sqrt{5}$……….(iii)

Now, $x^{4}+\frac{1}{x^{4}}= (x^{2}+\frac{1}{x^{2}})^{2}-2x^{2}\frac{1}{x^{2}}$

= $((x+\frac{1}{x})^{2}-2x\frac{1}{x})^{2}-2$

=$((2\sqrt{5})^{2}-2)^{2}-2$

=322

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