If x is a real number satisfying the equation x^2 + \frac{18}{x^2 + 3} = 8, what is the va… — GMAT

If $x$ is a real number satisfying the equation $x^2 + \frac{18}{x^2 + 3} = 8$ , what is the value of $x^2$ ?

Cevap: 6

The value of

x^2

is 6.

Substituting

u = x^2

converts the equation to

u + \frac{18}{u+3} = 8

. Multiplying through by

(u+3)

yields

u^2 + 3u + 18 = 8u + 24

, which simplifies to the quadratic equation

u^2 - 5u - 6 = 0

. Factoring gives

(u-6)(u+1) = 0

, so

u = 6

u = -1

. Because

x

must be a real number,

x^2

must be non-negative, eliminating

u = -1

. Thus,

x^2 = 6

Substitute a temporary variable

u = x^2

into the given equation

u + \frac{18}{u + 3} = 8

Using substitution simplifies the fourth-degree rational equation into a standard quadratic structure.

Clear the denominator by multiplying the entire equation by

(u + 3)

u(u + 3) + 18 = 8(u + 3)

, which expands to

u^2 + 3u + 18 = 8u + 24

Eliminating fractions allows rearranging terms into polynomial form.

Rearrange terms into standard quadratic form

au^2 + bu + c = 0

u^2 - 5u - 6 = 0

Moving all terms to one side prepares the expression for factoring.

Factor the quadratic polynomial into linear binomial factors

(u - 6)(u + 1) = 0

, giving

u = 6

u = -1

The roots of the transformed equation represent candidate values for

x^2

Apply the constraint that

x

is a real number to test candidate values

Since

x^2 \geq 0

for all real

x

u = -1

is rejected, leaving

u = x^2 = 6

Squaring any real number yields a non-negative result.

Solving rational equations reducible to quadratics via algebraic substitution and rejecting non-real roots.

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