A rational expression with one square subtracted from another often simplifies by factoring. For short multiple-choice options, an allowed-number test in Desmos is fast: pick values that keep the denominator nonzero, then evaluate the original and each option. One mismatch rules an option out. A match at one pair alone doesn't prove equivalence for every pair, so factoring explains why the result holds.
Hints
- Hint 1
An allowed input doesn't make the denominator zero. Try and , then check . If it's nonzero, you can compare every expression using that same pair.
- Hint 2
An expression equivalent for all allowed inputs must agree with the original at the pair you chose. Evaluate the fraction and then each option in Desmos. Which options can you rule out from a mismatch?
Step-by-step
Approach 1: Test an allowed pair in Desmos
- Step 1
Choose a pair the fraction allows
Try and . Type them as and so Desmos uses them as numbers rather than graph variables. Then type . Desmos shows , so the denominator is not zero and this pair is allowed.
- Step 2
Find the original fraction's value
Type the original fraction using and . Desmos shows . Any equivalent expression must give at this allowed pair.
- Step 3
Compare every option at the same pair
The first option is already on screen. Type the other three using and . Desmos shows , , , and , in that order. The first three fail at an allowed pair, so they cannot work for all allowed values. The only match is . Choice D.
Approach 2: Factor to prove the match
- Step 1
Recognize the two squares
The numerator is a difference of squares: is , and is . Write those squares explicitly:
- Step 2
Factor the numerator
A difference of squares follows . Here and , so the factors have opposite signs:
- Step 3
Put the factors over the original denominator
Replace the numerator with its factored form. The denominator now appears as a whole factor on top:
- Step 4
Cancel the matching factor
Because the original expression is defined only when , divide the top and bottom by that factor:
Cancel whole factors, not pieces of sums. The equivalent expression is for every allowed pair. Choice D.