Expert: Sherman D. - 3/26/2009

Question
QUESTION: I am having trouble on a test prep question. I have worked it a couple of times and cant seem to get the right answer. The problem is sq root of(2x+70)- sq root of (x+3)=1 Sorry for the funky wording I don't know to put the special characters into text

ANSWER: sqrt(2x + 70) - sqrt(x + 3) = 1

sqrt(2x + 70) = 1 + sqrt(x + 3)

square both sides

2x + 70 = (1 + sqrt(x + 3))(1 + sqrt(x + 3))

2x + 70 = 1 + sqrt(x + 3) + sqrt(x + 3) + (x + 3)

2x + 70 = x + 2sqrt(x + 3) + 4

x + 64 = 2sqrt(x + 3)

square both sides

(x + 64)(x + 64) = 4(x + 3)

x^2 + 64x + 64x + 4096 = 4x + 12

x^2 + 124 + 4084 = 0

using the quadratic formula

x = (-b +/- sqrt(b^2 - 4ac))/(2a)

x = (-124 +/- sqrt((124)^2 - 4(1)(4084)))/(2(1))

x = (-124 +/- sqrt(15376 - 16336))/2

x = (-124 +/- sqrt(-960))/2

x = (-124 +/- sqrt(-64 * 15))/2

x = (-124 +/- 8isqrt(15))/2

ANS : -62 +/- 4sqrt(15)i

so there is no real solution.

---------- FOLLOW-UP ----------

QUESTION: I am so sorry it was sqrt(2x+7) instead of sqrt (2x+70)

Answer
sqrt(2x + 7) - sqrt(x + 3) = 1

using the same exact method, only taking out the 0.

sqrt(2x + 7) = 1 + sqrt(x + 3)

square both sides

2x + 7 = 1 + 2sqrt(x + 3) + x + 3

x + 3 = 2sqrt(x + 3)

square both sides

x^2 + 6x + 9 = 4x + 12

x^2 + 2x - 3 = 0

(x + 3)(x - 1) = 0

x = -3 or 1

sometimes you have to treat the sqrts with negative answers, as in
-sqrt(2x + 7) instead of sqrt(2x + 7) when you plug in the -3 or 1, but not in this case, you can just leave the problem like it is.

www.quickmath.com will give you the answers but it won't show the work.