Problem source

Let $\mathrm{h}(x)=ax^{2}+bx+c,$ where $a,b$ and $c$ are constants, and $a\neq0$. Give a condition which $a,b$ and $c$ must satisfy in order that $\mathrm{h}(x)$ can be written in the form 
\[
a(x+k)^{2},\tag{*}
\]
where $k$ is a constant. 
If $\mathrm{f}(x)=3x^{2}+4x$ and $\mathrm{g}(x)=x^{2}-2$, find the two constant values of $\lambda$ such that $\mathrm{f}(x)+\lambda\mathrm{g}(x)$ can be written in the form $(*)$. Hence, or otherwise, find constants $A,B,C,D,m$ and $n$ such that 
\begin{alignat*}{1}
\mathrm{f}(x) & =A(x+m)^{2}+B(x+n)^{2}\\
\mathrm{g}(x) & =C(x+m)^{2}+D(x+n)^{2}.
\end{alignat*}
If $\mathrm{f}(x)=3x^{2}+4x$ and $\mathrm{g}(x)=x^{2}+\alpha$ and it is given by that there is only one value of $\lambda$ for which $\mathrm{f}(x)+\lambda\mathrm{g}(x)$ can be written in the form $(*)$, find $\alpha$.

Solution source

For $h(x)$ to be written in this form $b^2=4ac$.

Suppose $f(x) = 3x^2+4x$, $g(x) = x^2-2$. then,

\begin{align*}
&& f(x) + \lambda g(x) &= (3+\lambda)x^2+4x - 2 \lambda \\
\Rightarrow && 0 &= 16 + 8(3+\lambda) \lambda \\ 
\Rightarrow && 0 &= 2+ 3 \lambda + \lambda^2 \\
&&&= (\lambda +1)(\lambda + 2) \\
\Rightarrow && \lambda &= -1 , -2 \\
\end{align*}

\begin{align*}
&& f(x) - g(x) &= 2(x+1)^2 \\
&& f(x) -2g(x) &= (x+2)^2  \\
\Rightarrow && g(x) &= 2(x+1)^2 - (x+2)^2  \\
&& f(x) &= 4(x+1)^2 - (x+2)^2
\end{align*}

Suppose $f(x) = 3x^2+4x, g(x) = x^2 + \alpha$, then 

\begin{align*}
&& f(x) + \lambda g(x) &= (3+\lambda)x^2+4x+\lambda \alpha \\
\Rightarrow && 0 &= 16  -2\lambda \alpha(\lambda + 3) \\
&& 0 &= \alpha \lambda^2 +3\lambda-8 \\
 \Rightarrow && 0 &= 9 +32 \alpha \\
\Rightarrow && \alpha &= -\frac{9}{32}
\end{align*}