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Friday, 4 December 2015

QUESTION(Quadratic Equations) Solve the equation x^2+12x+36=64

QUESTION(Quadratic Equations)
.Solve the equation x^2+12x+36=64

SOLUTION: =================
I shall do this with two methods as I am not sure which you are familiar with. 

We have 

x^2 + 12x + 36 = 64
x^2 + 12x + 36 - 64 = 0
x^2 + 12x - 28 = 0

Method 1 Factorise to get   
(x - 2)(x + 14)=0

so, 
x - 2 = 0 and  x + 14 = 0
This gives x = 2, -14

Method 2 : Use the quadratic formula (see below in appendix) 
This gives 
  x = [ - 12 \pm \sqrt{12^2 - 4(1)(-28)} ] / (2(1))
 x = [ -12 \pm \sqrt{256} ] / 2
x = [ -12 \pm 16 ]/2
x = [-12 + 16]/2  ,\quad[-12-16]/2
x = 2, -14    

CHECK:  
when x = 2,
 LHS = x^2 + 12x + 36  
=2^2 +12(2) + 36
= 4 + 24 + 36
= 28 + 36
= 64 = RHS 
when x = -14
LHS = x^2 + 12x + 36
= (-14)^2 +12(-14) + 36
= 196 - 168 + 36
= 28 + 36
= 64 = RHS 


======================================================================= 
Appendix: The Quadratic Formula 

The quadratic equation ax^2 + bx + c = 0  has solutions given by 
                                  x ={ -b \pm \sqrt{b^2 - 4ac} \over 2a}

Real solutions only exist provided b^2 - 4ac \ge 0.

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