SOLUTION: (3+ √x)^2 + 3(3+√x)-10=0 u = (3+√x) u^2= (3+√x)^2 u^2 + 3u - 10 = 0 x= [-b ± √(b^2-4ac)] ÷ 2a -3 ± √(3^2-4(1)(-10)) ÷ (2)(1) -3 ± √
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-> SOLUTION: (3+ √x)^2 + 3(3+√x)-10=0 u = (3+√x) u^2= (3+√x)^2 u^2 + 3u - 10 = 0 x= [-b ± √(b^2-4ac)] ÷ 2a -3 ± √(3^2-4(1)(-10)) ÷ (2)(1) -3 ± √
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Question 313371
:
(3+ √x)^2 + 3(3+√x)-10=0
u = (3+√x)
u^2= (3+√x)^2
u^2 + 3u - 10 = 0
x= [-b ± √(b^2-4ac)] ÷ 2a
-3 ± √(3^2-4(1)(-10)) ÷ (2)(1)
-3 ± √(9 + 40 )÷ 2
-3 ± √(49 ) ÷ 2
-3 + 7 = 4/2 = 2
-3 – 7 = -10/2 = -5
What did I do wrong?
Answer by
Fombitz(32388)
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Show Source
):
You can
put this solution on YOUR website!
Nothing. It looks good.
But you're not done yet.
Remember you solved for u.
You should have u in the quadratic formula and not x.
Now find x using your original substitution.
