If x > 1 is an integer such that \sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}} = 6, … — GMAT

If $x > 1$ is an integer such that $\sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}} = 6$ , what is the value of $x$ ?

By rewriting the expressions under the radicals using the perfect square identity

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

, the equation simplifies directly to

(\sqrt{x-1} + 1) + (\sqrt{x-1} - 1) = 6

. Adding the terms gives

2\sqrt{x-1} = 6

, which leads to

\sqrt{x-1} = 3

, and squaring both sides yields

x - 1 = 9

, so

x = 10

Recognize the nested radical identity structure

Observe that

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

and

x - 2\sqrt{x-1} = (\sqrt{x-1} - 1)^2

Rewriting the terms under the outer square roots as perfect squares eliminates the outer radicals.

Simplify the square roots

Since

x > 1

\sqrt{x-1} + 1 > 0

and

\sqrt{x-1} - 1 \ge 0

for

x \ge 2

. Thus,

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

The principal square root of

a^2

|a|

, which is positive.

Set the simplified expression equal to 6 and solve for

x

2\sqrt{x-1} = 6 \implies \sqrt{x-1} = 3 \implies x - 1 = 9 \implies x = 10

Isolating the radical and squaring both sides yields the exact integer value of

x

Simplifying nested radicals using algebraic identities

Question