If x = \sqrt{11 + 4\sqrt{6}} - \sqrt{11 - 4\sqrt{6}} and \left(\frac{1}{16}\right)^{x^2 - … — State PSC Exam

If $x = \sqrt{11 + 4\sqrt{6}} - \sqrt{11 - 4\sqrt{6}}$ and $\left(\frac{1}{16}\right)^{x^2 - 15} = 64^{y + 1}$ , what is the value of $y$ ?

Answer: 1

The value of

y

1

Simplifying

\sqrt{11 \pm 4\sqrt{6}}

\sqrt{8} \pm \sqrt{3}

yields

x = 2\sqrt{3}

and

x^2 = 12

. Substituting

x^2

gives

(1/16)^{-3} = 16^3 = 2^{12}

. Matching exponents with

64^{y+1} = 2^{6y+6}

yields

12 = 6y + 6

, giving

y = 1

Simplify the nested surds for

x

x = 2\sqrt{3}

Express

11 \pm 4\sqrt{6}

11 \pm 2\sqrt{24} = (\sqrt{8} \pm \sqrt{3})^2

, giving

\sqrt{11 + 4\sqrt{6}} - \sqrt{11 - 4\sqrt{6}} = (2\sqrt{2} + \sqrt{3}) - (2\sqrt{2} - \sqrt{3}) = 2\sqrt{3}

Evaluate

x^2

x^2 = 12

Squaring

x = 2\sqrt{3}

yields

x^2 = 4 \times 3 = 12

Substitute

x^2 = 12

into the exponent of the given equation

\left(\frac{1}{16}\right)^{-3} = 64^{y + 1}

The exponent

x^2 - 15

evaluates to

12 - 15 = -3

Convert both sides to a common prime base (base 2)

2^{12} = 2^{6y + 6}

\left(\frac{1}{16}\right)^{-3} = 16^3 = (2^4)^3 = 2^{12}

and

64^{y + 1} = (2^6)^{y + 1} = 2^{6y + 6}

Equate powers and solve for

y

y = 1

Since the bases are identical,

12 = 6y + 6 \implies 6y = 6 \implies y = 1

Nested surd evaluation via perfect square conversion combined with solving exponential equations using index rules.

Question