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There exist constants a, h and k such that

x^2 + 12x + 4 = a(x - h)^2 + k

for all real numbers x.  Enter the ordered triple (a,h,k).

 Jul 13, 2021
 #1
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Answer: \((1,-6,-32)\)

 

Solution:

 

(x-h)² can be simplified into x² - 2hx + h². Multiplying that by a gives ax² - 2ahx + ah². Adding k gives ax² - 2ahx + ah² + k. This leads to the equation:

 

\(x^2+12x+4=ax^2-2ahx+ah^2+k\)

 

Looking at the x² term shows that a must be equal to 1. This turns the equation into this:

 

\(x^2+12x+4=x²-2hx+h^2+k\)

 

Since -2hx is the only term containing x¹ on the right side, and 12x is the only one on the left side containing x¹, -2hx=12x. This means that h is equal to -6. The equation turns into the following:

 

\(x^2+12x+4=x^2+12x+36+k\)

 

Using the same logic as the previous, 36+k=4. Solving this gives k = -32.

 

Plugging all of this information into an ordered triple (a, h, k) gives \((1,-6,-32)\).

 Jul 13, 2021

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