When x+frac{1}{x}=3,find x^{2}+frac{1}{x^{2}} | vedclass

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

Question

(D) Given that $x+\frac{1}{x}=3$.To find the value of $x^{2}+\frac{1}{x^{2}}$,we square the given equation on both sides:$\left(x+\frac{1}{x}\right)^{

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) Given that $x+\frac{1}{x}=3$.To find the value of $x^{2}+\frac{1}{x^{2}}$,we square the given equation on both sides:$\left(x+\frac{1}{x}\right)^{2} = 3^{2}$Using the algebraic identity $(a+b)^{2} = a^{2}+b^{2}+2ab$,we get:$x^{2} + \frac{1}{x^{2}} + 2(x)\left(\frac{1}{x}\right) = 9$$x^{2} + \frac{1}{x^{2}} + 2 = 9$Subtracting $2$ from both sides:$x^{2} + \frac{1}{x^{2}} = 9 - 2 = 7$Therefore,the value is $7$.

When $x+\frac{1}{x}=3$,find $x^{2}+\frac{1}{x^{2}}$.

Similar Questions

Solve the given two equations and select the correct answer from the given options.
$I.$ $17 x^{2} + 48 x = 9$
$II.$ $13 y^{2} = 32 y - 21$

For what value of $\lambda$ is the sum of the squares of the roots of ${x^2} + (2 + \lambda )x - \frac{1}{2}(1 + \lambda ) = 0$ minimum?

If $x+y+z=0,$ then find the value of $\frac{(x+y)(y+z)(z+x)}{x y z}$

The area of a rectangle is the same as that of a circle of radius $\frac{\sqrt{35}}{\sqrt{11}} \text{ cm}$. If the length of the rectangle exceeds its breadth by $3 \text{ cm}$,find the dimensions of the rectangle.

The number of solutions of $\frac{\log 5 + \log (x^2 + 1)}{\log (x - 2)} = 2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module