If sqrt{x}+frac{1}{sqrt{x}}=5, what will be t | vedclass

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

Question

(C) Given: $\sqrt{x}+\frac{1}{\sqrt{x}}=5$Squaring both sides:$\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2} = 5^{2}$$x + 2\left(\sqrt{x}\right)\left(\

Vedclass · Accepted Answer

$527$

Vedclass · Answer

(C) Given: $\sqrt{x}+\frac{1}{\sqrt{x}}=5$Squaring both sides:$\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2} = 5^{2}$$x + 2\left(\sqrt{x}\right)\left(\frac{1}{\sqrt{x}}\right) + \frac{1}{x} = 25$$x + 2 + \frac{1}{x} = 25$$x + \frac{1}{x} = 23$Now,squaring both sides again:$\left(x + \frac{1}{x}\right)^{2} = 23^{2}$$x^{2} + 2(x)\left(\frac{1}{x}\right) + \frac{1}{x^{2}} = 529$$x^{2} + 2 + \frac{1}{x^{2}} = 529$$x^{2} + \frac{1}{x^{2}} = 529 - 2 = 527$

If $\sqrt{x}+\frac{1}{\sqrt{x}}=5,$ what will be the value of $x^{2}+\frac{1}{x^{2}}$?

Similar Questions

Resolve into factors: $(a+b)^{2}-14 c(a+b)+49 c^{2}$.

Factorize: $a^{2}+\frac{1}{a^{2}}+3-2 a-\frac{2}{a}$

If $x = \sqrt[3]{x^{2} + 11} - 2$,then the value of $(x^{3} + 5x^{2} + 12x)$ is

Let $f(x) = a_{0}x^{n} + a_{1}x^{n-1} + a_{2}x^{n-2} + \ldots + a_{n-1}x + a_{n}$,where $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$ are constants. If $f(x)$ is divided by $ax - b$,then the remainder is:

If $x+\frac{1}{y}=1$ and $y+\frac{1}{z}=1,$ find the value of $z+\frac{1}{x}.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module