For all x > 0, which of the following is equivalent to the expression \frac{4x^2 - 25}{2x^… — SAT

For all $x > 0$ , which of the following is equivalent to the expression $\frac{4x^2 - 25}{2x^{\frac{3}{2}} + 5x^{\frac{1}{2}}}$ ?

The correct equivalent expression is

2x^{\frac{1}{2}} - 5x^{-\frac{1}{2}}

To simplify the expression

\frac{4x^2 - 25}{2x^{\frac{3}{2}} + 5x^{\frac{1}{2}}}

, we first factor the numerator and the denominator. The numerator is a difference of squares:

4x^2 - 25 = (2x - 5)(2x + 5)

. In the denominator, we can factor out

x^{\frac{1}{2}}

to get

x^{\frac{1}{2}}(2x + 5)

. Substituting these factored forms gives

\frac{(2x - 5)(2x + 5)}{x^{\frac{1}{2}}(2x + 5)}

. Canceling the common factor

(2x + 5)

yields

\frac{2x - 5}{x^{\frac{1}{2}}}

. Dividing each term in the numerator by the denominator gives

\frac{2x}{x^{\frac{1}{2}}} - \frac{5}{x^{\frac{1}{2}}}

. Applying exponent rules, this simplifies to

2x^{1 - \frac{1}{2}} - 5x^{-\frac{1}{2}} = 2x^{\frac{1}{2}} - 5x^{-\frac{1}{2}}

, which is the correct expression.

Factor the numerator of the expression as a difference of squares.

4x^2 - 25 = (2x - 5)(2x + 5)

To simplify the rational expression, we need to factor both the numerator and the denominator to identify common factors.

Factor out the common term

x^{\frac{1}{2}}

from the denominator.

2x^{\frac{3}{2}} + 5x^{\frac{1}{2}} = x^{\frac{1}{2}}(2x + 5)

Factoring out

x^{\frac{1}{2}}

reveals the common binomial factor

(2x + 5)

in the denominator.

Substitute the factored forms back into the original expression and cancel the common factor

(2x + 5)

\frac{(2x - 5)(2x + 5)}{x^{\frac{1}{2}}(2x + 5)} = \frac{2x - 5}{x^{\frac{1}{2}}}

Since

x > 0

, the term

2x + 5

is non-zero, allowing us to cancel it from both the numerator and the denominator.

Divide each term in the numerator by

x^{\frac{1}{2}}

and apply the rules of exponents.

\frac{2x}{x^{\frac{1}{2}}} - \frac{5}{x^{\frac{1}{2}}} = 2x^{1 - \frac{1}{2}} - 5x^{-\frac{1}{2}} = 2x^{\frac{1}{2}} - 5x^{-\frac{1}{2}}

This simplifies the rational expression to its final equivalent form.

Equivalent Algebraic Expressions

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