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I am having trouble figuring out: 9x^2 + 18x + 5 =0

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2 Answers

 
  f(x) =9X2 +18X + 5
 
    First check factorability of the quadratic,  find out whether there exist 2 integers , a, b, such that
    their Sum equals 18, and their product is 9*5 =45
    To do that :
     Write 45 as product of its factors:  45 = 3 * 15 = 9 * 5
      Observe that 3+ 15 =18
 
     f(X) = 9X2 +15X  + 3X +5                    /Break up 18X =   15X +3X
                3X(3X + 5 ) +(3x +5)              / factor the quadratic by grouping
               ( 3X + 5 )( 3X + 1) =0
                     3X + 5 = 0       X = -5/3
              3X + 1 = 0       X = -1/3
       In the event if, in quadratic aX2 + bX +c
       there is no 2 integers , m;n, does not exist such that m.n= ac and m+n = b
      Then to solve the quadratic you can either use factoring by completing square or
       using quadratic formula.   
    
Hi William;
9x2 + 18x + 5 =0
For the FOIL...
FIRST must be (3x)(3x) or (9x)(x).
OUTER and INNER must add-up to 18x.
LAST must be (5)(1) or (1)(5).
(3x+5)(3x+1)=0
FOIL...
FIRST...(3x)(3x)=9x2
OUTER...(3x)(1)=3x
INNER...(5)(3x)=15x
LAST...(5)(1)=
9x2+3x+15x+5=0
9x2+18x+5=0
So...
(3x+5)(3x+1)=0
One or both of the parenthetical equations must equal zero.
3x+5=0
3x=-5
x=-5/3
and/or
3x+1=0
3x=-1
x=-1/3