PreCalculus
posted by Joseph P. .
I think I need help with this problem
3+ square root(x6) = square root of (x+9)
I thought I should square everything to get rid of square root signs
so I had 9 + x6 = x+ 9
then I would have x+ 3once I subtracted 96 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

3 + √(x6) = √(x+9)
you have to square the whole side, not just each term
9 + 6√(x6) + x6 = x+9
6√(x6) = 6
√(x6) = 1
square again
x6 = 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 (x6) come from and where did the 9 disappear to

left side squared
=(3+√(x6) )^2
= (3+√(x6))(3+√(x6))
= 9 + 3√(x6) + 3√(x6) + x6
= 9 + 6√(x6) + x6
right side squared = √(x+9) ^2 = x+9
so
9 + 6√(x6) + x6 = x+9 , collect all terms to right side, leave 6√.... on the left
6√(x6) = xx + 99 + 6
6√(x6) = 6
divide both sides by 6
√(x6) = 1
square again
x6 = 1
etc