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If \(x^2\)\(-3x+2 = (x-k)^2+p\), what is the value of \(p\)?

a.) None of the below

\(b.) 2\)

\(c.) -2\)

\(d.) -1/4\)

\(e.) 1 \)

 Apr 29, 2022

Best Answer 

 #1
avatar+147 
+3

Classic complete the square problem ^_^

If you expand the right side, we have

\(x^2 - 3x + 2 = x^2 - 2kx + k^2 + p\).

Because like terms have to match on both sides, we garner \(-3x = -2kx \implies k = \frac{3}{2}\).

Now, we need \(2 = k^2 + p.\) Substituting the value for \(k\) we just found, we get \(2 = \left(\frac{3}{2}\right)^2 + p \implies 2 = \frac{9}{4} \implies p = -1/4\).

 

So, we can conclude the answer \(d.) -1/4\).

 Apr 29, 2022
 #1
avatar+147 
+3
Best Answer

Classic complete the square problem ^_^

If you expand the right side, we have

\(x^2 - 3x + 2 = x^2 - 2kx + k^2 + p\).

Because like terms have to match on both sides, we garner \(-3x = -2kx \implies k = \frac{3}{2}\).

Now, we need \(2 = k^2 + p.\) Substituting the value for \(k\) we just found, we get \(2 = \left(\frac{3}{2}\right)^2 + p \implies 2 = \frac{9}{4} \implies p = -1/4\).

 

So, we can conclude the answer \(d.) -1/4\).

Anthrax Apr 29, 2022
 #2
avatar+122385 
+1

Nice, Anthrax  !!!!

 

 

cool cool cool

CPhill  Apr 29, 2022

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