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What is the sum of all possible values of for which x^2 + kx - 10x + 36 + 12x - 8 is the square of a binomial?

 Feb 16, 2022
 #1
avatar+122390 
+1

x^2  + kx -10x + 36 + 12x  - 8     combine like terms

 

x^2 + (12 - 10 + k)x + 28

 

x^2  + ( 2 + k)x  + 28

 

If this can be written as the square  of a binomial  the discrminant  =  0

 

So

 

(2 + k)^2  - 4 (1) (28)   = 0

 

(2 + k)^2  =  112

 

2 + k  =  ±√112   = ± 4√ 7

 

k =  ± 4√7   - 2

 

The sum of all possible k's =   (4√ 7  - 2)  + ( -4 √ 7 -2 )  =  -4 

 

 

cool cool cool

 Feb 16, 2022
 #2
avatar+36417 
+1

From CPhill answer :

     x^2  + ( 2 + k)x  + 28     

 

    the roots are the same    R1 = R2 

       From Vieta                   

          R1R1 = 28              so   R1 = +- sqrt 28

                     and        -  ( R1 + R1) /a  =      -R1 - R1  =  2 + k                   (a = 1 in this case)

                                                                    - 2R1 = 2+k

                                                                      - 2(+- sqrt28) - 2   = k

                                                                        - 4 sqrt7  -2 = k    and  4 sqrt 28 -2 = k        summed = -4

                                                                                                                      (as Chris found via different method)

 Feb 16, 2022

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