For all positive real numbers a and b, the custom operation \diamondsuit is defined by a \… — GRE General Test

For all positive real numbers $a$ and $b$ , the custom operation $\diamondsuit$ is defined by $a \diamondsuit b = \frac{a^2 + b^2}{ab}$ . The function $f$ is defined for all $x > 0$ by $f(x) = x \diamondsuit 4$ . If $f(x) = 2.5$ , what is the value of $x$ that is greater than $4$ ?

Answer: 8

Answer

The value of

x

greater than

4

8

Applying the custom operator gives

f(x) = \frac{x^2 + 16}{4x}

. Setting this equal to

2.5

yields

\frac{x^2 + 16}{4x} = \frac{5}{2}

, which simplifies to

x^2 - 10x + 16 = 0

. The roots are

x = 2

and

x = 8

. Since

x

must be greater than

4

, the only valid answer is

8

Step-by-Step Solution

Substitute

a = x

and

b = 4

into the custom operation definition

a \diamondsuit b = \frac{a^2 + b^2}{ab}

f(x) = \frac{x^2 + 16}{4x}

This establishes the explicit algebraic rule for the function

f(x)

Set

f(x)

equal to

2.5

and clear the fraction.

\frac{x^2 + 16}{4x} = 2.5 \implies x^2 + 16 = 10x

Multiplying both sides by

4x

converts the rational equation into a standard polynomial equation.

Rearrange into standard quadratic form and solve by factoring.

x^2 - 10x + 16 = 0 \implies (x - 2)(x - 8) = 0 \implies x = 2

x = 8

Factoring determines all potential positive real solutions for

x

Select the solution satisfying the constraint

x > 4

x = 8

The question explicitly specifies that

x

must be greater than

4

, eliminating

x = 2

Key Concept

Evaluating custom binary operations and solving algebraic function equations involving quadratic constraints.

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