If x is a real number that satisfies the equation \sqrt{2x + 7} + \sqrt{x + 3} = 1, what i… — ACT

If $x$ is a real number that satisfies the equation $\sqrt{2x + 7} + \sqrt{x + 3} = 1$ , what is the value of $x$ ?

Answer: -3

The only real solution to the equation is

-3

The value

-3

is the only real number that satisfies the original equation. Substituting

-3

back into the original equation yields

\sqrt{2(-3) + 7} + \sqrt{-3 + 3} = \sqrt{1} + 0 = 1

, which is true.

Isolate the first radical term.

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

This allows for squaring both sides to eliminate one radical.

Square both sides and simplify.

2x + 7 = x + 4 - 2\sqrt{x + 3}

Squaring removes the radical on the left side, though it creates a middle term on the right side.

Isolate the remaining radical term.

x + 3 = -2\sqrt{x + 3}

Grouping the non-radical terms on one side prepares the equation for a second squaring step.

Square both sides again to eliminate the remaining radical.

x^2 + 6x + 9 = 4(x + 3)

Squaring both sides eliminates the radical completely, converting the expression into a polynomial equation.

Solve the quadratic equation.

x = -3

and

x = 1

Rearranging to

x^2 + 2x - 3 = 0

and factoring as

(x + 3)(x - 1) = 0

gives the candidate solutions.

Substitute candidates back into the original equation to check for extraneous solutions.

The only valid solution is

x = -3

Substituting

x = 1

yields

5 = 1

(invalid), while substituting

x = -3

yields

1 = 1

(valid).

Solving radical equations by isolating radicals and squaring, then testing for extraneous solutions.

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