The cue for all allowed values means these two rational expressions, fractions of polynomials, agree at every permitted input. First, find the inputs that make the original denominator zero. Then use a Desmos list regression at two legal inputs to find the two unknown constants. Two matching inputs alone wouldn't prove equivalence, but the question already promises it.
Hints
- Hint 1
A denominator of makes a fraction undefined. Factor the original denominator before choosing inputs. Even if a factor cancels later, its zero is still forbidden in the original expression.
- Hint 2
The problem gives an identity, an equality that holds at every allowed input. Since there are two unknown constants, try two different allowed inputs, such as and , to get information about both.
- Hint 3
A Desmos regression fits unknown constants to the inputs you give it. Put and in a list named , then replace every in the given equality with and change to .
Step-by-step
Approach 1: Fit the two constants in Desmos
Step 1Find the forbidden inputs
The denominator is the bottom of the fraction, and it can't be . Factor the original one:
So , , and are forbidden. In particular, don't use merely because the expression on the right is defined there.
- Step 2
Choose two allowed inputs
Type into Desmos; it shows the two entries in the list. Neither input is forbidden. Because equality is already promised for every allowed input, two legal inputs can pin down the two unknown constants. The next line will use both entries.
- Step 3
Read the constant the question asks for
Replace each with and the equality sign with so Desmos fits and to both inputs. Type:
Under PARAMETERS, Desmos shows and . The question asks for , not , so the value of is . Choice C.
Approach 2: Match the constant terms
Step 1Cancel the shared factor
Factor out of the top and bottom, then cancel that whole factor. This works only where :
- Step 2
Expose the target denominator
Factor the remaining denominator to show the from the target form:
So the given equality becomes
- Step 3
Clear the denominators
For allowed , multiply both sides by . The on the right contributes the entire product, while the fraction contributes :
- Step 4
Match the parts without x
This is an identity, so its constant terms, the parts without , must agree. The left contributes . On the right, contributes , and contributes :
- Step 5
Solve for k
Add to both sides:
Divide by :
So the constant in the requested form is . Choice C.
Lessons that teach this
- SAT Advanced AlgebraIntermediateCoreRewrite rational expressions and preserve restrictions
- SAT Advanced AlgebraAdvancedUse polynomial identities, factors, and unknown coefficients
- DesmosAdvancedCoreFind constants in equivalent expressions
- DesmosIntermediateCoreRestrictions, piecewise functions, and rational expressions