Posted by **Joseph P.** on Wednesday, September 21, 2011 at 7:28pm.

I think I need help with this problem-

3+ square root(x-6) = square root of (x+9)

I thought I should square everything to get rid of square root signs

so I had 9 + x-6 = x+ 9

then I would have x+ 3once I subtracted 9-6 from the left side = x+9 on the right side which doesn't make any sense to move anything anymore it doesn't work what did I do wrong? Thanks

- Pre-Calculus -
**Reiny**, Wednesday, September 21, 2011 at 7:33pm
3 + √(x-6) = √(x+9)

you have to square the whole side, not just each term

9 + 6√(x-6) + x-6 = x+9

6√(x-6) = 6

√(x-6) = 1

square again

x-6 = 1

x = 7

check: This is MUST, since we squared each side

LS = 3 + √1 = 4

RS = √16 = 4

therefore x = 7

- Pre-Calculus-Please recheck -
**Joseph P**, Wednesday, September 21, 2011 at 7:41pm
I'm probably just not understanding yet but Where did the 6 in front of square root of (x-6) come from and where did the 9 disappear to

- Pre-Calculus -
**Reiny**, Wednesday, September 21, 2011 at 7:58pm
left side squared

=(3+√(x-6) )^2

= (3+√(x-6))(3+√(x-6))

= 9 + 3√(x-6) + 3√(x-6) + x-6

= 9 + 6√(x-6) + x-6

right side squared = √(x+9) ^2 = x+9

so

9 + 6√(x-6) + x-6 = x+9 , collect all terms to right side, leave 6√.... on the left

6√(x-6) = x-x + 9-9 + 6

6√(x-6) = 6

divide both sides by 6

√(x-6) = 1

square again

x-6 = 1

etc

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