Question

If $x^4+\frac{1}{x^4}=34$, then the value of $(x-\frac{1}{x})^2$ will be

• A. 7
• B. -7
• C. -3
• D. 4

Answer 1

Answer:

D. 4

Solution:

$x^4+\frac{1}{x^4}=34$

⇒ adding 2 on both side we get

$⇒x^4+(\frac{1}{x^4}+2=34+2$

$⇒(x^2)^2+(\frac{1}{x^2})^2+2\times{x^2}\times{\frac{1}{x^2}}=36$

⇒$(x^2+\frac{1}{x^2})^2=36$

⇒$(x^2+\frac{1}{x^2})=6$     -----1

⇒ we need to find the value of $(x-\frac{1}{x})^2$

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

⇒ put the value from equation 1 in above equation we will get

$(x-\frac{1}{x})^2=6-2$

∴ $(x-\frac{1}{x})^2=4$