Question

Solve: $(x^2-3x{)}^2-16(x^2-3x)-36=0$.

Answer 1

We have, $(x^2-3x{)}^2-16(x^2-3x)-36=0$

Substitute $y$ for $x^2-3x$ in the given equation, we get
$y^2-16y-36=0$.

Let's solve this equation using factorisation method

$\begin{array}{cc}{y}^2-16y-36& =0\\ y^2-18y+2y-36& =0\\ y(y-18)+2(y-18)& =0\\ (y-18)(y+2)& =0\\ \Rightarrow y-18& =0\text{or}y+2=0\\ \Rightarrow y& =18\text{or}y=-2\end{array}$

Since $x^2-3x=y$, thus for $y=18$, $x$ can be calculated as,

$\begin{array}{cc}{x}^2-3x& =18\\ x^2-3x-18& =0\\ x^2-6x+3x-18& =0\\ x(x-6)+3(x-6)& =0\\ (x-6)(x+3)& =0\\ \Rightarrow x-6& =0\text{or}x+3=0\\ \Rightarrow x& =6\text{or}x=-3\end{array}$

Now, for $y=-2$, we have
$\begin{array}{cc}{x}^2-3x& =-2\\ x^2-3x+2& =0\\ x^2-x-2x+2& =0\\ x(x-1)-2(x-1)& =0\\ (x-1)(x-2)& =0\\ \Rightarrow x-1& =0\text{or}x-2=0\\ \Rightarrow x& =1\text{or}x=2\end{array}$

$\therefore$ Required solution is $x=1,2,6,-3$.