If x^{4}+frac{1}{x^{4}}=119 and x1,then find | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $x^{4}+\frac{1}{x^{4}}=119$.Adding $2$ to both sides,we get $x^{4}+\frac{1}{x^{4}}+2=119+2=121$.This can be written as $(x^{2}+\frac{1}{x^{2

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(C) Given $x^{4}+\frac{1}{x^{4}}=119$.Adding $2$ to both sides,we get $x^{4}+\frac{1}{x^{4}}+2=119+2=121$.This can be written as $(x^{2}+\frac{1}{x^{2}})^{2}=11^{2}$.Taking the square root,$x^{2}+\frac{1}{x^{2}}=11$ (since $x>1$,$x^{2}+\frac{1}{x^{2}}$ must be positive).Now,subtract $2$ from both sides: $x^{2}+\frac{1}{x^{2}}-2=11-2=9$.This simplifies to $(x-\frac{1}{x})^{2}=3^{2}$,so $x-\frac{1}{x}=3$.To find $x^{3}-\frac{1}{x^{3}}$,we use the identity $(x-\frac{1}{x})^{3} = x^{3}-\frac{1}{x^{3}}-3(x-\frac{1}{x})$.Substituting the value $x-\frac{1}{x}=3$,we get $3^{3} = x^{3}-\frac{1}{x^{3}}-3(3)$.$27 = x^{3}-\frac{1}{x^{3}}-9$.Therefore,$x^{3}-\frac{1}{x^{3}} = 27+9 = 36$.

If $x^{4}+\frac{1}{x^{4}}=119$ and $x>1$,then find the positive value of $x^{3}-\frac{1}{x^{3}}$.

Similar Questions

$I. 24x^2 + 38x + 15 = 0$
$II. 12y^2 + 28y + 15 = 0$

Solve the given two equations and select the correct answer from the given options.
$I.$ $\frac{12 \times 4}{x^{4/7}} - \frac{3 \times 4}{x^{4/7}} = x^{10/7}$
$II.$ $y^3 + 783 = 999$

The values of $a$ for which $2x^2 - 2(2a + 1)x + a(a + 1) = 0$ may have one root less than $a$ and other root greater than $a$ are given by

If $a+b+c=0,$ then find the value of $a^{3}+b^{3}+c^{3}$.

Find the sum of all integral values of $a$ such that the equation $||x - 2| - |3 - x|| = 2 - a$ has a solution.

Vedclass Test Series

Exam Paper Generator

Online Exam Module