# Pre-Calculus

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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 - ,

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

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 - ,

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