Q13. Extraneous Solutions

 x - a  = x - 4

If a = 2, what is the solution set of the equation above?

A) {3, 6}

B) {2}

C) {3}

D) {6}

They tell you a = 2 so

 x - 2  = x - 4

Now you square each side

x-2 = x2 -8x + 16

x = x2 -8x + 18

x2 -9x + 18= 0

From here you’d factor

(x-3)(x-6) = 0

So x = {3,6}

BUT…. whenever you have a root or possible division by 0, you should always go back and check your work.

If you go back and plug 3, you’ll find it doesn’t work

3-2 = 3-4

1 <> -1

Check 6 and you’ll find it works

6-2 = 6-4

2 = 2

So the only solution is D.

Lesson

There are some questions where it actually may make sense to plug in answers rather than solving it. Actually, some questions require that approach. In the above, I actually solved the problem and then tested out the answers.

The important thing here is that you need to go plug your answers when you have a radical or potential division by zero question. The SAT loves to throw in these types of odd ball questions.

Want more practice questions? Check out the Worthington Prep SAT Tutoring Questions page.

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Q14. Elimination

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Q12. Line of Best Fit