If $x=\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}$, then determine the value of $x^2+3x-15$?

The correct answer is $49+26\sqrt{6}$.

Let us solve the given problem step-by-step.

Step 1: Simplify the given expression for $x$.
We are given:

$x=\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}$

To simplify the expression under the outer square root, rationalise the denominator by multiplying the numerator and denominator inside the square root by $5+2\sqrt{6}$:

$x=\sqrt{\frac{(5+2\sqrt{6})(5+2\sqrt{6})}{(5-2\sqrt{6})(5+2\sqrt{6})}}$

Using the algebraic identity $(a-b)(a+b)=a^2-b^2$ in the denominator:

$5^2-{\left(2\sqrt{6}\right)}^2=25-24=1$

So, the expression inside the square root simplifies to:

$x=\sqrt{{(5+2\sqrt{6})}^2}$

$x=5+2\sqrt{6}$

Step 2: Find $x^2$.
Squaring both sides of $x=5+2\sqrt{6}$:

$x^2={(5+2\sqrt{6})}^2$

Using the algebraic identity ${(a+b)}^2=a^2+2ab+b^2$:

$x^2=5^2+2\left(5\right)\left(2\sqrt{6}\right)+{\left(2\sqrt{6}\right)}^2$

$x^2=25+20\sqrt{6}+24$

$x^2=49+20\sqrt{6}$

Step 3: Calculate the value of $x^2+3x-15$.
Substitute the values of $x$ and $x^2$ into the expression:

$x^2+3x-15=(49+20\sqrt{6})+3(5+2\sqrt{6})-15$

Expand the expression:

$=49+20\sqrt{6}+15+6\sqrt{6}-15$

Combine like terms:

$=49+(15-15)+(20\sqrt{6}+6\sqrt{6})$

$=49+26\sqrt{6}$

Thus, the value of $x^2+3x-15$ is $49+26\sqrt{6}$.