Question

Given that $2x=3+\sqrt{7}$

Find the value of $4x^2+\frac{1}{x^2}$

• A. 343/729
• B. 729/343
• C. 49/81
• D. 81/49

Answer 1

The correct option is A

343/729

Given that
$2x=3+\sqrt{7}$

Hence $x=\frac{3+\sqrt{7}}{2}$

$x^2=(\frac{3+\sqrt{7}}{2}{)}^2$

= $\frac{9+6\sqrt{7}+7}{4}$ ( expanding using $(a+b{)}^2=a^2+2ab+b^2$

$4x^2$ = $\frac{16+6\sqrt{7}}{4}$ ---------------------------------(i)

$x^2=(\frac{8+3\sqrt{7}}{2}$) ( Taking 4 common and simplifying)

$\frac{1}{x^2}=\frac{2}{8+3\sqrt{7}}$

Multiplying Numerator and Denominator with the rationalising factor of Denominator,we get

$\frac{1}{x^2}$ = $\frac{2}{8+3\sqrt{7}}$ x $\frac{8-3\sqrt{7}}{8-3\sqrt{7}}$

$\frac{1}{x^2}$ = $\frac{2(8-3\sqrt{7})}{(8{)}^2-(3\sqrt{7}{)}^2}$

$\frac{1}{x^2}$ = $\frac{16-6\sqrt{7}}{64-63}$

$\frac{1}{x^2}$ =$16-6\sqrt{7}$ ______________________(ii)

Adding (i) and (ii)
$4x^2+\frac{1}{x^2}=\left(\frac{4(16+6\sqrt{7}}{4}\right)+16-6\sqrt{7}$

= $16+6\sqrt{7}+16-6\sqrt{7}$

= 16 + 16
= 32