Solving for $p$ and $k$.

Question

For the question,

$$F(x)= x^{2} + 6x + 20 + k(x^{2} -3x -12)$$

I have to find the value of k and the value of $p$ given that the minimum value of $F(x)$ is $p$ and $F(-2) = p$.

What I did is: I expanded the expression of $F(x)$ and reached a term for $p= 12 - 2p$. I tried to complete the square of the two quadratics of $F(x)$ separately then added the two terms while also correcting for coefficients. I doubt this operation however went for it anyway due to any alternatives seem wrong.

I reached the term $-\frac {39}4k + 11$ then promptly equated it to $12 - 2k$.

I reached the value of $-\frac{4}{31}$ for $K$.

My textbook has answers for $k = \frac{2}{7}$ and $p=\frac{80}{7}$.

Answer 1

You can try completing the square directly. Notice that the minimum of $F(x)$ lies at $x = -2$, and $F(-2) = p$. This means that

$$F(x) = a(x + 2)^2 + p$$

for some constant $a$. We can't assume $a = 1$ because any non-zero value of $a$ can work, so we leave it there as it is.

Now we go ahead and solve as usual by comparing coefficients:

$$a(x + 2)^2 + p = x^2 + 6x + 20 + k(x^2 - 3x - 12)$$ $$ax^2 + 4ax + (4a + p) = (k + 1)x^2 + (6 - 3k)x + (20 - 12k)$$

Comparing coefficients, we have

$$k + 1 = a~~~~,~~~~4a = 6 - 3k~~~~,~~~~4a + p = 20 - 12k$$ Firstly, $$4a = 6 - 3(a - 1)$$ $$4a = 9 - 3a$$ $$7a = 9$$ $$a = \frac{9}{7}$$ Secondly, $$\begin{align}k &= a - 1 \\&= \frac{9}{7} - 1 \\&= \frac{2}{7}\end{align}$$ And lastly, $$4a + p = 20 - 12k \implies 4\cdot\frac{9}{7} + p = 20 - 12 \cdot\frac{2}{7}$$ $$p = \frac{80}{7}$$

Answer 2

A quadratic $ax^2+bx+c$ with a positive leading term $a\gt0$ attains its minimum at $-\cfrac{b}{2a}\,$.

Rewriting $F(x) = (1+k)x^2+(6-3k)x+20-12k\,$ it follows that: $$-\cfrac{6-3k}{2(1+k)}=-2 \quad \implies \quad k = \cfrac{2}{7}$$

(Since the leading term $1+k = 1 + \cfrac{2}{7}\gt 0$ this is a valid solution.)

Substituting $k$ back gives $F(x)=\cfrac{9}{7} x^2+\cfrac{36}{7} x + \cfrac{116}{7}\,$, then $p=F(-2)=\cfrac{80}{7}\,$.