If x is a real number that satisfies the equation x - \sqrt{x + 2} = 4, what is the value … — GRE General Test

If $x$ is a real number that satisfies the equation $x - \sqrt{x + 2} = 4$ , what is the value of $(x + 2)^{\frac{3}{2}}$ ?

$27$ Cevap
B
$8$
C
$7\sqrt{7} + 2\sqrt{2}$
D
$-27$
E
$81$

Cevap

Isolating the radical gives

x - 4 = \sqrt{x + 2}

. Squaring both sides yields

(x - 4)^2 = x + 2

, which expands to

x^2 - 8x + 16 = x + 2

, or

x^2 - 9x + 14 = 0

. Factoring yields

(x - 7)(x - 2) = 0

, giving solutions

x = 7

and

x = 2

. Testing

x = 7

in the original equation gives

7 - \sqrt{9} = 4

, which is valid. Testing

x = 2

gives

2 - \sqrt{4} = 0 \neq 4

, which is extraneous. Substituting the valid root

x = 7

into

(x + 2)^{\frac{3}{2}}

gives

(7 + 2)^{\frac{3}{2}} = 9^{\frac{3}{2}} = (\sqrt{9})^3 = 3^3 = 27

Adım Adım Çözüm

Isolate the radical term in the given equation.

x - 4 = \sqrt{x + 2}

Isolating the radical allows both sides to be squared cleanly.

Square both sides to eliminate the square root and form a quadratic equation.

(x - 4)^2 = x + 2 \implies x^2 - 8x + 16 = x + 2 \implies x^2 - 9x + 14 = 0

Squaring eliminates the radical and allows standard quadratic solving techniques.

Factor the quadratic equation to find candidate solutions.

(x - 7)(x - 2) = 0 \implies x = 7 \text{ or } x = 2

Factoring provides potential real roots.

Check candidate solutions in the original equation to eliminate extraneous roots.

For

x = 7

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

(valid).
For

x = 2

2 - \sqrt{2 + 2} = 2 - 2 = 0 \neq 4

(extraneous).

Squaring both sides can introduce false solutions that fail the original radical equation.

Evaluate the target expression

(x + 2)^{\frac{3}{2}}

using

x = 7

(7 + 2)^{\frac{3}{2}} = 9^{\frac{3}{2}} = (9^{\frac{1}{2}})^3 = 3^3 = 27

Applying exponent rules (

a^{\frac{m}{n}} = (\sqrt[n]{a})^m

) gives the exact required value.

Anahtar Kavram

Solving radical equations requires isolating the radical, squaring both sides, checking for extraneous solutions introduced by squaring, and evaluating fractional exponents via roots and integer powers.

Tahmini Süre:2m 0s

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