Posted by Anonymous on Tuesday, April 26, 2011 at 2:54pm.
how do i solve this completing the square problem: 3x^2+12x+10=0

trig  MathMate, Tuesday, April 26, 2011 at 3:31pm
3x^2+12x+10=0
3(x^2+4x+10/3)=0
(x+2)²4+10/3=0
(x+2)²(√(2/3))²=0
(x+2+√(2/3))(x+2√(2/3))=0
therefore
x=2√(2/3) or 2+√(2/3)

trig  bobpursley, Tuesday, April 26, 2011 at 3:34pm
divide by 3
x^2+4x+10/3=0
x^2+4x + ???= 10/3 + ???
well, look at the second term. take one half of it, and square it, add to both sides.
x^2+4x+4=10/3+4= 10/3+12/3
(x+2)^2= 2/3
take the square root of each side.
x+2= +sqrt 2/3
x=2+sqrt 2/3
check that.
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