Question 134

If $$\ x^{2}\ - 3x + 1 = 0$$, then what is the value of $$\ x^{4}\ $$+ $$\ \frac{1}{x^{4}}$$?

Solution

Given : $$x^2-3x+1=0$$

Dividing both sides by $$'x'$$

=> $$x+\frac{1}{x}=3$$

Squaring both sides, we get :

=> $$x^2+\frac{1}{x^2}+2(x)(\frac{1}{x})=9$$

=> $$x^2+\frac{1}{x^2}=9-2=7$$

Again squaring both sides,

=> $$x^4+\frac{1}{x^4}+2(x^2)(\frac{1}{x^2})=49$$

=> $$x^4+\frac{1}{x^4}=49-2=47$$

=> Ans - (C)

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