GATE2018 CH: GA-9

Question

If $x^{2}+ x - 1 = 0$ what is the value of $x^4 + \dfrac{1}{x^4}$?

• A. $1$
• B. $5$
• C. $7$
• D. $9$

Answer 1

$\left ( x^{2}+x-1 \right )=0$

$\implies\left ( x^{2}-1 \right )=-x$

$\implies\left ( x^{2}-1 \right )^{2}=\left ( -x \right )^{2}$

$\implies x^{4}-2x^{2}+1=x^{2}$

$\implies x^{4}+1=3x^{2}$

$\implies\left ( x^{4}+1 \right )^{2}=\left ( 3x^{2} \right )^{2}$

$\implies x^{8}+2x^{4}+1=9x^{4}$

$\implies x^{8}+1=7x^{4}$

Now, we can divide  both sides by $x^{4}$ and get

$x^{4}+\frac{1}{x^{4}}=7$

Comments

@srestha mam,

Can you refer any books for solving this kind of polynomial questions?

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If you got the logic Do the practice more.

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Ok bro. thanks.

Answer 2

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

Divide both side by $'x'.$

$\implies \dfrac{x^{2} + x - 1}{x} = \dfrac{0}{x}\:\:, x\neq 0$

$\implies  x + 1 - \dfrac{1}{x} = 0$

$\implies x-\dfrac{1}{x} = -1$

Take square both side

$\implies \left(x-\dfrac{1}{x}\right)^{2} = (-1)^{2}$

$\implies x^{2} + \dfrac{1}{x^{2}} - 2 = 1$

$\implies x^{2} + \dfrac{1}{x^{2}}  = 3$

Again take square both side

$\implies \left(x^{2}+\dfrac{1}{x^{2}}\right)^{2} = 3^{2}$

$\implies x^{4}+\dfrac{1}{x^{4}}+ 2 = 3^{2}$

$\implies x^{4}+\dfrac{1}{x^{4}} = 7$

So, the correct answer is $(C).$