If \sqrt{x} + \frac{1}{\sqrt{x}} = 3, x > 0, then x^2(x^2 - 47) = ?

Question 110

If $$\sqrt{x} + \frac{1}{\sqrt{x}} = 3, x > 0$$, then $$x^2(x^2 - 47) = ?$$

Solution

Expression : $$\sqrt{x} + \frac{1}{\sqrt{x}} = 3, x > 0$$ ---------------(i)

Squaring both sides,

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

=> $$x+\frac{1}{x}=7$$

Again squaring both sides,

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

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

=> $$x^4-47x^2+1=0$$

=> $$x^2(x^2-47)=-1$$

=> Ans - (D)

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