If \(x^2\)\(-3x+2 = (x-k)^2+p\), what is the value of \(p\)?
a.) None of the below
\(b.) 2\)
\(c.) -2\)
\(d.) -1/4\)
\(e.) 1 \)
+150 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 \(d.) -1/4\).
+150 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 \(d.) -1/4\).
