If x is a real number satisfying the radical equation \sqrt{x + 7} - x = 1, what is the va… — GRE General Test

If $x$ is a real number satisfying the radical equation $\sqrt{x + 7} - x = 1$ , what is the value of $x$ ?

The value of

x

2

Isolating the square root yields

\sqrt{x + 7} = x + 1

. Squaring both sides produces

x + 7 = x^2 + 2x + 1

, which reduces to the quadratic equation

x^2 + x - 6 = 0

. Factoring yields

(x + 3)(x - 2) = 0

, giving potential roots of

x = -3

and

x = 2

. Testing

x = 2

in the original expression yields

\sqrt{2+7} - 2 = 3 - 2 = 1

, confirming it is correct.

Isolate the radical expression on one side of the equation.

\sqrt{x + 7} = x + 1

Isolating the radical allows squaring both sides cleanly to eliminate the radical sign.

Square both sides of the equation.

x + 7 = (x + 1)^2 = x^2 + 2x + 1

Squaring removes the square root on the left side.

Rearrange into standard quadratic form

ax^2 + bx + c = 0

x^2 + x - 6 = 0

Subtracting

x + 7

from both sides collects all terms on one side.

Factor the quadratic equation and solve for potential values of

x

(x + 3)(x - 2) = 0 \implies x = -3 \text{ or } x = 2

Setting each factor to zero identifies potential solutions.

Substitute each potential solution back into the original equation

\sqrt{x + 7} - x = 1

to check for extraneous roots.

For

x = 2

\sqrt{2 + 7} - 2 = 3 - 2 = 1

(Valid). For

x = -3

\sqrt{-3 + 7} - (-3) = 2 + 3 = 5 \neq 1

(Extraneous).

Squaring an equation can introduce false solutions that must be eliminated.

Solving radical equations and checking for extraneous solutions

Tahmini Süre:1m 30s

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