Answer:
C. Add a radical term to both sides and square both sides twice
Step-by-step explanation:
We suppose you want to solve ...
[tex]\sqrt{b+20}-\sqrt{d}=5\\\\\sqrt{b+20}=5+\sqrt{d}\qquad\text{add a radical term}\\\\b+20=25+10\sqrt{d}+d\qquad\text{square both sides once}\\\\b-d-5=10\sqrt{d}\qquad\text{subtract right-side non-radical terms}\\\\(b-d-5)^2=100d\qquad\text{square both sides a second time}[/tex]
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We note that the solution summary (choice C) doesn't mention the fact that you need to separate the remaining radical term from the others after the first squaring.
The result is a quadratic function that produces extraneous solutions. This is shown in the attached graph. The red function is the original relation between b (x) and d (y). The blue function includes the red function and another branch that is all extraneous solutions.
