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05132013  #1 
Join Date: Dec 2011
Posts: 386

Quadratic Equations
Find the solution set for these:
This one is the hardest....
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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; 05132013 at 06:08 PM.. 
05192013  #2  
Join Date: Dec 2008
Posts: 249

Quote:
(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) 

09172014  #3 
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^24ac)/2a 
09192014  #4  
Join Date: Dec 2011
Posts: 386

Quote:
Hey Mr.Hui can you tell me where the sqared option is?
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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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