Question

Could I have some help with extraneous problems, please?

Answer 1

@jdoe0001 @johnweldon1993

Answer 2

Mr. Jones is asked to write two types of radical equations for the SAT Prep Course. The first solution to the radical equation must be extraneous. The second solution to the radical equation must be non-extraneous. Write one equation where the solution is extraneous. Then write a second equation where the solution is non-extraneous.

Answer 3

Hmm...if I gave you the equation

\[\large \sqrt{x} = x - 6\]

solve that for me :)

Answer 4

First step? Square both sides right?

\[\large \sqrt{x}^2 = (x - 6)^2\]
\[\large x = (x - 6)^2\]

Second step? expand that right side out

\[\large x = x^2 - 12x + 36\]

Third? Subtract x from both sides of the equation
\[\large x^2 - 13x + 36 = 0\]

can you solve that quadratic for me? should get 2 easy answers if I did it right :)

Answer 5

Oh sorry! Does x=9?

Answer 6

That's 1 answer....what about the second? :)

Answer 7

4?

Answer 8

Perfect..so if x = 9 and x = 4

Lets check to make sure they both work

If we check 9..

\[\large \sqrt{x} = x -6\]
\[\large \sqrt{9} = 9 -6\]
\[\large 3 = 3\]

So that works..check 4 for me :)

Answer 9

2=-2

Answer 10

Ohh looky there...it doesnt work out...thus we have...and extraneous solution :)

Answer 11

Fabulous! How about a non extraneous solution?

Answer 12

Well a non extraneous ...hmm try

\[\large \sqrt{x - 2} = 5\]

Answer 13

x=27!

Answer 14

Mmhmm :) and that's the ONLY solution...no extraneous thing trying to confuse us >.< lol :P

Answer 15

Okay, got it. Thanks again! Really appreciate you OpenStudy people!

Answer 16

Lol no problem :P it's what we're here for :)

Answer 17

Would you mind helping me with one last problem?