Question
how do i solve this completing the square problem: 3x^2+12x+10=0
Answers
MathMate
3x^2+12x+10=0
3(x^2+4x+10/3)=0
(x+2)²-4+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)
3(x^2+4x+10/3)=0
(x+2)²-4+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)
bobpursley
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.
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.