What is the sum of all real solutions to the equation x - 3 = \sqrt{x + 3}? — GRE General Test

What is the sum of all real solutions to the equation $x - 3 = \sqrt{x + 3}$ ?

The sum of all real solutions to the equation is 6.

Squaring both sides of

x - 3 = \sqrt{x + 3}

gives

x^2 - 6x + 9 = x + 3

, which simplifies to

x^2 - 7x + 6 = 0

. Factoring yields

(x - 6)(x - 1) = 0

, giving candidate solutions

x = 6

and

x = 1

. Substituting

x = 6

into the original equation gives

3 = \sqrt{9} = 3

, which is valid. Substituting

x = 1

gives

-2 = \sqrt{4} = 2

, which is false, making

x = 1

an extraneous solution. Therefore,

x = 6

is the single valid solution, and its sum is 6.

Eliminate the radical by squaring both sides of the equation.

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

Squaring both sides converts the radical equation into a standard quadratic equation.

Rearrange the terms into standard quadratic form

ax^2 + bx + c = 0

and factor.

x^2 - 7x + 6 = 0 \implies (x - 6)(x - 1) = 0

Subtracting

x

and

3

from both sides isolates zero on one side so the quadratic expression can be factored.

Solve for potential values of

x

x = 6

x = 1

Setting each factor to zero yields the algebraic solutions to the quadratic equation.

Check each potential solution in the original radical equation

x - 3 = \sqrt{x + 3}

For

x = 6

6 - 3 = 3

and

\sqrt{6 + 3} = 3

(Valid). For

x = 1

1 - 3 = -2

and

\sqrt{1 + 3} = 2

(Invalid, since

-2 \neq 2

Squaring an equation can introduce extraneous solutions that do not satisfy the original principal square root relationship.

Calculate the sum of all valid real solutions.

The only valid real solution is

6

, so the sum is

6

Extraneous solutions must be discarded before computing the final sum.

Solving radical equations using quadratic factoring and identifying extraneous solutions

Estimated Time:2m 0s

Question