If , which of the following is equivalent to

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SAT Equivalent Expressions Question 75 | Free SAT Prep | Aniko

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Question 75·Hard·Equivalent Expressions

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If $k>0$ , which of the following is equivalent to

\frac{6k^{-\frac12}-2k^{\frac32}}{2k^{\frac12}}\cdot\frac{k^{-1}-k^{2}}{k^{2}}

For expression-equivalence questions with exponents, first simplify the given expression yourself: factor out common terms, use exponent rules ( $k^a/k^b = k^{a-b}$ and $k^{-n} = 1/k^n$ ), and cancel where possible before looking at the choices. Once you have a clean form—often a single fraction—compare its structure (especially signs in factors like $3 - k^2$ vs. $k^2 - 3$ ) to the answer choices instead of trying to expand everything. If you’re unsure, you can also plug in a simple allowed value (like $k=1$ or $k=2$ , avoiding values that make denominators zero) into the original expression and each choice to quickly rule out incorrect options.

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Separate the two fractions

Before multiplying, try simplifying each fraction on its own. Focus first on $\frac{6k^{-1/2}-2k^{3/2}}{2k^{1/2}}$ , then on $\frac{k^{-1}-k^{2}}{k^{2}}$ .

Use exponent rules in the first fraction

In the first fraction, factor out the greatest common factor from the numerator. What common numerical and $k$ -power factor do $6k^{-1/2}$ and $2k^{3/2}$ share, and how does dividing powers of $k$ work ( $k^a/k^b = k^{a-b}$ )?

Simplify the second fraction term by term

For $\frac{k^{-1}-k^{2}}{k^{2}}$ , divide each term in the numerator by $k^{2}$ . Use the exponent rule $k^{a}/k^{b} = k^{a-b}$ and then rewrite any negative exponents as fractions.

After multiplying, look at factor signs

Once you multiply the simplified fractions, you should get something like $(3 - k^2)(1 - k^3)$ over a power of $k$ . How can you factor out $-1$ from each of these to rewrite them in the more standard $(k^2 - \text{constant})$ and $(k^3 - \text{constant})$ forms?

Desmos Guide

Enter the original expression as a function

In Desmos, type something like

f(x) = \frac{6x^{-1/2}-2x^{3/2}}{2x^{1/2}} \cdot \frac{x^{-1}-x^{2}}{x^{2}}

so you can evaluate or graph the original expression for positive $x$ (remember the problem states $k>0$ , so only look at $x>0$ ).

Enter each answer choice as separate functions

Define functions for each option, for example:

$A(x) = \frac{(x^2 - 3)(x^3 - 1)}{x^4}$
$B(x) = \frac{(3 - x^2)(x^3 - 1)}{x^4}$
$C(x) = x - \frac{3}{x} + \frac{1}{x^2} - \frac{3}{x^4}$
$D(x) = \frac{(x^2 - 3)(x^3 + 1)}{x^4}$ These represent the four answer choices in terms of $x$ .

Compare values or graphs for positive inputs

Use Desmos’s table feature (click the gear icon next to each function and add a table) or visually inspect the graphs for $x>0$ . Pick a few positive $x$ -values (like $x=1, 2, 3$ ) and compare $f(x)$ to each option’s value. The option whose function always matches $f(x)$ for these positive values is the equivalent expression.

Step-by-step Explanation

Simplify the first fraction

Work with

\frac{6k^{-1/2}-2k^{3/2}}{2k^{1/2}}.

Factor $2$ from the numerator:

6k^{-1/2}-2k^{3/2}=2(3k^{-1/2}-k^{3/2}).

Next factor out $k^{-1/2}$ from the parentheses. Since $k^{3/2} = k^{-1/2}k^2$ :

3k^{-1/2}-k^{3/2} = k^{-1/2}(3 - k^2).

So the numerator is $2k^{-1/2}(3 - k^2)$ , and the whole fraction becomes

\begin{aligned} \frac{2k^{-1/2}(3 - k^2)}{2k^{1/2}} &= (3 - k^2)\cdot \frac{k^{-1/2}}{k^{1/2}} \\ &= (3 - k^2)\cdot k^{-1} \\ &= \frac{3 - k^2}{k}. \end{aligned}

Simplify the second fraction

Now simplify

\frac{k^{-1} - k^{2}}{k^{2}}.

Divide each term in the numerator by $k^2$ :

\begin{aligned} \frac{k^{-1} - k^{2}}{k^{2}} &= \frac{k^{-1}}{k^{2}} - \frac{k^{2}}{k^{2}} \\ &= k^{-3} - 1. \end{aligned}

Rewrite with positive exponents and a single denominator:

\begin{aligned} k^{-3} - 1 &= \frac{1}{k^{3}} - 1 \\ &= \frac{1 - k^{3}}{k^{3}}. \end{aligned}

Multiply the two simplified fractions

Now multiply the results from Steps 1 and 2:

\frac{3 - k^{2}}{k} \cdot \frac{1 - k^{3}}{k^{3}}.

Multiply the numerators and denominators:

\begin{aligned} \frac{3 - k^{2}}{k} \cdot \frac{1 - k^{3}}{k^{3}} &= \frac{(3 - k^{2})(1 - k^{3})}{k \cdot k^{3}} \\ &= \frac{(3 - k^{2})(1 - k^{3})}{k^{4}}. \end{aligned}

This is a simplified, factored form of the original expression, but we can rewrite the factors in a more standard order.

Rewrite factors to match an answer choice

Rewrite each factor so that the variable term comes first.

Note that

3 - k^{2} = -(k^{2} - 3), \quad 1 - k^{3} = -(k^{3} - 1).

So the numerator becomes

\begin{aligned} (3 - k^{2})(1 - k^{3}) &= [-(k^{2} - 3)]\cdot [-(k^{3} - 1)] \\ &= (k^{2} - 3)(k^{3} - 1) \end{aligned}

(because the product of two negatives is positive).

Therefore, the original expression is equivalent to

\frac{(k^{2} - 3)(k^{3} - 1)}{k^{4}}

which corresponds to choice A.

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