Question

If $\sqrt{\frac{2+\sqrt3}{2-\sqrt3}}$, then determine the value of the $x^2+x-9$ ?

• A. 3√3
• B. 5√3
• C. 7√3
• D. 9√3

Answer 1

Answer:

B. 5√3

First simplify:

$x = \sqrt{\frac{2+\sqrt3}{2-\sqrt3}}$

Rationalize inside:

$\frac{2+\sqrt3}{2-\sqrt3} \times \frac{2+\sqrt3}{2+\sqrt3}$

$= \frac{(2+\sqrt3)^2}{4-3}$

$(2+\sqrt3)^2 = 4+4\sqrt3+3 = 7+4\sqrt3$

So,

$x = \sqrt{7+4\sqrt3}$

Now observe:

$(2+\sqrt3)^2 = 7+4\sqrt3$
$\Rightarrow x = 2+\sqrt3$

Now compute:

$x^2 + x - 9$

First:

$x^2 = (2+\sqrt3)^2 = 7+4\sqrt3$

So,

$x^2 + x - 9 = (7+4\sqrt3) + (2+\sqrt3) - 9$
$7+4\sqrt3+2+\sqrt3-9 = (9-9) + 5\sqrt3 = 5\sqrt3$