If x is a real number that satisfies the equation \sqrt{x + 6\sqrt{x - 9}} + \sqrt{x - 6\s… — GRE General Test

If $x$ is a real number that satisfies the equation $\sqrt{x + 6\sqrt{x - 9}} + \sqrt{x - 6\sqrt{x - 9}} = 10$ , what is the value of $x$ ?

Answer: 34

Using the substitution

u = \sqrt{x - 9} \ge 0

, we have

x = u^2 + 9

. The expressions under the square roots become

x + 6\sqrt{x - 9} = u^2 + 6u + 9 = (u + 3)^2

and

x - 6\sqrt{x - 9} = u^2 - 6u + 9 = (u - 3)^2

. Taking square roots gives

\sqrt{(u + 3)^2} + \sqrt{(u - 3)^2} = (u + 3) + |u - 3| = 10

. For

u \ge 3

, this simplifies to

(u + 3) + (u - 3) = 10 \implies 2u = 10 \implies u = 5

. Finally, substituting

u = 5

back yields

x = 5^2 + 9 = 34

Define a variable substitution to simplify the nested radical structure.

Let

u = \sqrt{x - 9}

where

u \ge 0

. Squaring both sides gives

u^2 = x - 9

, so

x = u^2 + 9

This substitution allows the expressions inside the outer square roots to be rewritten as polynomials in terms of

u

Rewrite the expressions under each square root as perfect square trinomials.

x + 6\sqrt{x - 9} = (u^2 + 9) + 6u = (u + 3)^2

and

x - 6\sqrt{x - 9} = (u^2 + 9) - 6u = (u - 3)^2

Expressing terms as perfect squares allows the outer radicals to be simplified.

Simplify the square root expressions using absolute values.

\sqrt{(u + 3)^2} + \sqrt{(u - 3)^2} = (u + 3) + |u - 3| = 10

For any real number

a

\sqrt{a^2} = |a|

. Since

u \ge 0

u + 3 > 0

, so

|u + 3| = u + 3

Solve the absolute value equation across valid domain intervals.

u \ge 3

|u - 3| = u - 3

, giving

(u + 3) + (u - 3) = 10 \implies 2u = 10 \implies u = 5

. If

0 \le u < 3

|u - 3| = 3 - u

, giving

(u + 3) + (3 - u) = 6 \neq 10

(no solution). Thus,

u = 5

Splitting into cases based on the definition of absolute value isolates the valid root.

Substitute

u = 5

back into the expression for

x

x = 5^2 + 9 = 25 + 9 = 34

Converting from

u

back to

x

provides the solution to the original equation.

Simplifying nested radicals by completing the square under the radical sign and applying the identity

\sqrt{a^2} = |a|

Question