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05-19-2013   #2
MAS1

Join Date: Dec 2008
Posts: 249

Quote:
 Originally Posted by Dragonballful23 Find the solution set for these: ${{x^2}+2x+1}=0$ ${{3x^2}+4x+1}=0$ ${{x^2}+3x}=-2$ This one is the hardest.... ${{2x^2}+10x}=-15$
x^2 + 2x + 1 = 0
(x + 1)(x + 1) = 0
x = -1

3x^2 + 4x + 1 = 0
(3x + 1)(x + 1) = 0
3x +1 = 0
3x = -1
x = -1/3
x + 1 = 0
x = -1
So x = -1/3, -1

x^2 + 3x = -2
x^2 + 3x +2 = 0
(x + 1)(x + 2) = 0
x = -1, -2

2x^2 + 10x = -15
x^2 + 5x = -15/2
x^2 + 5x + 25/4 = -15/2 + 25/4
(x + 5/2)^2 = -5/4
sqrt((x + 5/2)^2) = sqrt(-5/4)
x + 5/2 = + or - i(sqrt(5)/2)
x = -5/2 + or - i(sqrt(5)/2)