If x = \sqrt{6 + 2\sqrt{5}} - \sqrt{6 - 2\sqrt{5}} and \left(\frac{1}{27}\right)^{x - 5} =… — State PSC Exam

If $x = \sqrt{6 + 2\sqrt{5}} - \sqrt{6 - 2\sqrt{5}}$ and $\left(\frac{1}{27}\right)^{x - 5} = 9^{y + 1}$ , what is the value of $x + y$ ?

The value of

x + y

\frac{11}{2}

Simplifying

\sqrt{6 + 2\sqrt{5}}

yields

\sqrt{5} + 1

and

\sqrt{6 - 2\sqrt{5}}

yields

\sqrt{5} - 1

. Taking their difference gives

x = 2

. Substituting

x = 2

into the exponential equation gives

(1/27)^{-3} = 27^3 = (3^3)^3 = 3^9

. The right side is

9^{y+1} = (3^2)^{y+1} = 3^{2y+2}

. Equating exponents gives

2y + 2 = 9

, which leads to

y = 7/2

. Adding

x

and

y

gives

2 + 7/2 = 11/2

Simplify the nested surds for

x

\sqrt{6 + 2\sqrt{5}} = \sqrt{5} + 1

and

\sqrt{6 - 2\sqrt{5}} = \sqrt{5} - 1

For a nested surd

\sqrt{a \pm 2\sqrt{b}}

, express

a

m+n

and

b

m \cdot n

. Here

5+1=6

and

5 \cdot 1 = 5

Compute the value of

x

x = (\sqrt{5} + 1) - (\sqrt{5} - 1) = 2

Subtracting the second simplified surd from the first cancels out

\sqrt{5}

and adds the integer terms.

Express the exponential equation using a common prime base

3

(3^{-3})^{2 - 5} = (3^2)^{y + 1} \implies (3^{-3})^{-3} = 3^{2y + 2} \implies 3^9 = 3^{2y + 2}

Rewriting

\frac{1}{27} = 3^{-3}

and

9 = 3^2

allows direct comparison of exponents.

Solve for

y

and evaluate

x + y

2y + 2 = 9 \implies y = \frac{7}{2}

, so

x + y = 2 + \frac{7}{2} = \frac{11}{2}

Equating exponents with identical bases and adding the resolved values of

x

and

y

Simplification of nested surds

\sqrt{a \pm 2\sqrt{b}}

combined with solving exponential equations using common prime bases.

Tahmini Süre:2m 0s

Soru