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Old 05-13-2013   #1
Dragonballful23

 
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Default Quadratic Equations

Find the solution set for these:






This one is the hardest....

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Last edited by Mr. Hui; 05-13-2013 at 06:08 PM..
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Old 05-19-2013   #2
MAS1

 
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Quote:
Originally Posted by Dragonballful23 View Post
Find the solution set for these:






This one is the hardest....

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)
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Old 09-17-2014   #3
john123
 
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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
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Old 09-19-2014   #4
Dragonballful23

 
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Quote:
Originally Posted by john123 View Post
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?
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