Hi Mike, I’m asking about SAT 8, Section 3, Number 7. Is it always best to immediately plug in answer choices on questions like this? For the algebraic practice, I rewrote the equation as a quadratic and solved for (x=5) and (x= -1). Then sub’d each value back into the given equation to find that only (x=5) worked. OK…but what a time-killer at #7 out of 20. Any other solution path to consider? Thanks!

For questions like this (extraneous solution quadratics) the fastest solution IS to backsolve from the answer choices. If you look at the choices right away, you see that the only possible answers are 0, –1, and 5. Check those three out and you’re done in no time. Solve algebraically and you have to check the answers you get anyway to make sure they’re not extraneous!

To drive it home, here’s all the work you need to do if you backsolve:

    \begin{align*}\sqrt{2(0)+6}+4&=0+3\\\sqrt{6}+4&=3\end{align*}

    \begin{align*}\sqrt{2(-1)+6}+4&=-1+3\\\sqrt{4}+4&=2\\2+4&=2\\6&=2\end{align*}

    \begin{align*}\sqrt{2(5)+6}+4&=5+3\\\sqrt{16}+4&=8\\4+4&=8\\8&=8\end{align*}

Obviously only the last one of those worked out correctly, so the solution set contains only 5.

If you solve algebraically, as you point out, you eventually end up at –1 and 5 as possible answers (work below for anyone still reading), but because you had to square everything to get there, you need to check for extraneous solutions by doing the step you could have started with if you just backsolved from the beginning!

    \begin{align*}\sqrt{2x+6}+4&=x+3\\\sqrt{2x+6}&=x-1\\2x+6&=x^2-2x+1\\0&=x^2-4x-5\\0&=(x-5)(x+1)\end{align*}

 

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