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Find the sum of all values of $k$ for which $x^2 + kx - 9x + 25 - 9$ is the square of a binomial.

 Oct 31, 2024
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Find the sum of all values of $k$ for which $x^2 + kx - 9x + 25 - 9$ is the square of a binomial.   

 

          x^2 + kx - 9x + 25 - 9    ==>>    x2 + (kx – 9x) + 16   

 

          I combined like terms and arranged the expression in standard form  ax2 + bx + c    

          Note that (kx – 9x) can be written as (k – 9) x ... and (k – 9) will be our "b" value    

          To be a square, the "b" coefficient in the middle will be twice the square root of "c"    

 

          The sqrt(16) is + 4, therefore "b" will be either + 8 or – 8    

 

          For "b" = + 8    then (k – 9) = + 8    which means k = + 17    

          For "b" = – 8    then (k – 9) = – 8    which means k =   + 1    

 

                                          the sum of both values of k is    + 18    

 

          The problem doesn't ask for the binomial(s) but they're (x + 4)2 and (x – 4)2        

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 Oct 31, 2024

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