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 05-13-2013 #1 Dragonballful23     Join Date: Dec 2011 Posts: 386 Quadratic Equations 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.... ${{2x^2}+10x}=-15$ __________________ Finally finished all my tests!!! I am ekko main now ^_^ (ign: supersaiyan2363) It's not how much time you have, it's how you use it. -Ekko Last edited by Mr. Hui; 05-13-2013 at 06:08 PM..
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)

 09-17-2014 #3 john123   Join Date: Sep 2014 Posts: 4 To find roots of any quadratic equation , one formula is there if ax^2+bx+c is the given quadratic equation then roots are given by x= ( -b±√b^2-4ac)/2a
09-19-2014   #4
Dragonballful23

Join Date: Dec 2011
Posts: 386

Quote:
 Originally Posted by john123 To find roots of any quadratic equation , one formula is there if ax^2+bx+c is the given quadratic equation then roots are given by x= ( -b±√b^2-4ac)/2a
There is also a shortcut(my teachers tell me not use ) -b/2 plus/minus square root of (b/2)^2-ac

Hey Mr.Hui can you tell me where the sqared option is?
__________________
Finally finished all my tests!!!

I am ekko main now ^_^

(ign: supersaiyan2363)

It's not how much time you have, it's how you use it.
-Ekko

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