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Find the sum of the constants a, h, and k such that
2x^2 - 8x + 7 = a(x - h)^2 + k
for all real numbers x.

Respuesta :

sqdancefan

Answer:

  3

Step-by-step explanation:

The value of "a" is the coefficient of x^2, so we know that is 2.

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Solve for h

Now, we have ...

  2x^2 -8x +7 = 2(x -h)^2 +k

Expanding the right side gives us ...

  = 2(x^2 -2hx +h^2) +k

  = 2x^2 -4hx +2h^2 +k

Comparing x-terms, we see ...

  -4hx = -8x

  h = (-8x)/(-4x) = 2

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Solve for k

Now, we're left with ...

  2h^2 +k = 7 = 2(2^2) +k = 8 +k

Subtracting 8 we find k to be ...

  k = 7 -8 = -1

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And the sum of constants a, h, and k is ...

  a +h +k = 2 +2 -1 = 3

The sum of the constants is 3.

MindTrickery

Answer:

3

Step-by-step explanation:

The problem is to complete the square on the quadratic expression 2x^2 - 8x + 7. First, we write

                                    2x^2 - 8x + 7 = 2(x^2 - 4x) + 7

We want a square that includes the terms x^2 and -4x. This desired square is

                                      (x - 2)^2 = x^2 - 4x + 4

Hence,

                                    2(x^2 - 4x) + 7 &= 2[(x^2 - 4x + 4) - 4] + 7  

                                        = 2[(x - 2)^2 - 4] + 7  

                                        = 2(x - 2)^2 - 8 + 7  

                                        = 2(x - 2)^2 - 1.

Therefore, a = 2, h = 2, and k = -1, and a+h+k = 3

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