If x = 7 - 4sqrt{3},then find the value of x^ | vedclass

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

Question

(D) Given $x = 7 - 4\sqrt{3}$.First,find $\frac{1}{x}$:$\frac{1}{x} = \frac{1}{7 - 4\sqrt{3}}$.Rationalizing the denominator:$\frac{1}{x} = \frac{1}{7

Vedclass · Accepted Answer

$194$

Vedclass · Answer

(D) Given $x = 7 - 4\sqrt{3}$.First,find $\frac{1}{x}$:$\frac{1}{x} = \frac{1}{7 - 4\sqrt{3}}$.Rationalizing the denominator:$\frac{1}{x} = \frac{1}{7 - 4\sqrt{3}} 	imes \frac{7 + 4\sqrt{3}}{7 + 4\sqrt{3}} = \frac{7 + 4\sqrt{3}}{49 - (16 	imes 3)} = \frac{7 + 4\sqrt{3}}{49 - 48} = 7 + 4\sqrt{3}$.Now,find $x + \frac{1}{x}$:$x + \frac{1}{x} = (7 - 4\sqrt{3}) + (7 + 4\sqrt{3}) = 14$.We know that $(x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x^{2}} + 2$.Therefore,$x^{2} + \frac{1}{x^{2}} = (x + \frac{1}{x})^{2} - 2$.Substituting the value:$x^{2} + \frac{1}{x^{2}} = (14)^{2} - 2 = 196 - 2 = 194$.

If $x = 7 - 4\sqrt{3}$,then find the value of $x^{2} + \frac{1}{x^{2}}$.

Similar Questions

Find the value of $(625)^{-\frac{3}{4}}$.

Simplify: $\frac{8^{\frac{1}{3}} \times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}$

Fill in the blank to make the following statement true:
$(729)^{\frac{1}{3}} = \ldots$

Visualise $-4.126$ on the number line,using successive magnification.

$\frac{1}{\sqrt{9}-\sqrt{8}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module