For all real values of x, what is the real solution to the equation \sqrt{x+13} - 1 = x? — ACT

For all real values of $x$ , what is the real solution to the equation $\sqrt{x+13} - 1 = x$ ?

The only real solution to the equation is 3.

Isolating the radical term gives

\sqrt{x+13} = x+1

. Squaring both sides yields

x+13 = x^2 + 2x + 1

. Rearranging into standard quadratic form gives

x^2 + x - 12 = 0

, which factors as

(x+4)(x-3) = 0

. This yields candidate solutions

x = 3

and

x = -4

. Checking these values in the original equation shows that

x = 3

is a valid solution because

\sqrt{3+13} - 1 = 3

, whereas

x = -4

is extraneous because

\sqrt{-4+13} - 1 = 2 \neq -4

Isolate the radical expression on one side of the equation.

\sqrt{x+13} = x+1

This prepares the equation for squaring both sides to eliminate the radical.

Square both sides of the equation.

x+13 = (x+1)^2 \implies x+13 = x^2 + 2x + 1

Squaring both sides eliminates the radical and yields a polynomial equation.

Move all terms to one side to set the quadratic equation equal to zero.

x^2 + x - 12 = 0

This puts the equation in the standard quadratic form

ax^2 + bx + c = 0

so it can be solved by factoring.

Factor the quadratic equation.

(x+4)(x-3) = 0 \implies x = 3

x = -4

Factoring allows us to find the potential roots of the quadratic equation.

Check both candidate solutions in the original equation to identify any extraneous solutions.

For

x = 3

\sqrt{3+13} - 1 = 4 - 1 = 3

(valid). For

x = -4

\sqrt{-4+13} - 1 = 3 - 1 = 2 \neq -4

(extraneous).

Squaring both sides of an equation can introduce extraneous roots that must be checked and discarded.

Solving radical equations and checking for extraneous solutions

Tahmini Süre:1m 30s

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