Fundamental Drill#5

Given that $\begin{cases} f(x) = x^2+3x-4 = (x-1)(x+4) \\ g(x) = x^2-4x+4 = (x-2)^2 \end{cases}$

$\implies h(x) = f(x) + \alpha g(x) = (x-1)(x+4) + \alpha (x-2)^2 = p(x-q)^2$

Then we have

$\begin{aligned} h(1) & = 0 + \alpha = p(1-q)^2 & ...(1) \\ h(-4) & = 0 + 36\alpha = p(-4-q)^2 & ...(2) \\ h(2) & = 6 = p(2-q)^2 & ...(3) \end{aligned}$

$\begin{aligned} \frac {(2)}{(1)}: \quad \frac {(4+q)^2}{(1=q)^2} & = 36 \\ \frac {4+q}{1-q} & = 6 \\ 4+q & = 6 - 6q \\ \implies q & = \frac 27 \end{aligned}$

$\begin{aligned} (3): \quad p\left(2-\frac 27\right)^2 & = 6 \\ \implies p & = \frac {49}{24} \end{aligned}$

$\begin{aligned} (1): \quad \alpha & = \frac {49}{24}\left(1-\frac 27\right)^2 = \frac {25}{24} \end{aligned}$

Therefore, $m+n=25+24= \boxed{49}$ .

Relevant wiki: Quadratic Discriminant

$f(x)=x^2+3x-4$

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

$h(x)=f(x)+\alpha g(x)=x^2+3x-4+\alpha(x^2-4x+4)=(1+\alpha)x^2+(3-4\alpha)x+(4\alpha-4)$

When $h(x)=0$

$p(x-q)^2=0$

$x=q$

There is one real root only.

Or in other words, there are two equal real roots.

When $h(x)=0 \Rightarrow (1+\alpha)x^2+(3-4\alpha)x+(4\alpha-4)=0$

Discriminant $=(3-4\alpha)^2-4(1+\alpha)(4\alpha-4)=0$

$9-24\alpha+16\alpha^2-16(\alpha+1)(\alpha-1)=0$

$9-24\alpha+16\alpha^2-16(\alpha^2-1)=0$

$9-24\alpha+16=0$

$24\alpha=25$

$\alpha=\cfrac{25}{24}$

$p+q=25+24=49$

$\textbf{Alternative approach:}$ You can solve this too by completing the square as the quadratic formula is derived using completing the square method.