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For what constant k is 1 the minimum value of the quadratic 3x^2 - 15x + k over all real values of x? (x cannot be nonreal.)

 Oct 25, 2017
 #1
avatar+17746 
+1

For what constant k is 1 the minimum value of the quadratic 3x2 - 15x + k?

 

I'm going to find the value of k that will make the quadratic equal to zero and, then, add 1 to get the answer to the question.

 

Completing the square of:  3x2 - 15x

        Factor out the 3:          3(x2 - 5x)

        Finding the term that will complete the square:  divide 5 by 2 and square that answer:  (5/2)2  =  25/4  =  6 1/4

This means that   3x2 - 15x + 6 1/4   =   3(x - 5/2)2      is a perfect square

and, if you replace x with 2.5, your answer will be zero.

 

Since you want an answer of 1, you will need to add 1:  3(x - 5/2)2 + 1     =     3x2 - 15x + 18 3/4 + 1     =    3x2 - 15x + 19 3/4

 

Therefore, let  k = 19 3/4

 Oct 25, 2017
 #2
avatar+98197 
+1

Thanks, geno........here's another way

 

Tne x value that minimizes the function is given by  

 

-b / [2a ]       where  a = 3  and  b = -15     .....so we have

 

- [ -15] / [  2* 3 ]  =   15/6   =  5/2

 

So......putting this into the function we have

 

3 (5/2)^2 - 15 (5/2)  +   k  = 1 

 

75/4 -75/2 + k  = 1

 

-75/4  + k =  1

 

k =  79/4   = 19 + 3/4

 

 

cool cool cool

 Oct 25, 2017

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