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

Question

I cannot figure out how to get x from this problem. Thanks a lot!

Answer 1

 

  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.   

    

Answer 2

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