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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.)

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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.)

WhichWitchIsWhich  Oct 25, 2017
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#1
+17655
+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

geno3141  Oct 25, 2017
#2
+78744
+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

CPhill  Oct 25, 2017

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