Quadratics

1. Write $x^2+x=12$ as $x^2-x-12=0$ and then factorise to $(x-4)(x+3)=0$. The two factors $x-4$ and $x+3$ could both be zero to give a product of 0 and so the solutions are $x=4$ or $x=-3$.
2. First write $2x^2=12x-16$  as $2x^2-12x+16=0$ and then divide both sides by 2 to give $x^2-6x+8=0$. Then factorise to $(x-4)(x-2)=0$ and so the solutions are $x=4$ and $x=2$.

Since the question says find the solutions to two decimal places, it suggests that we should use the quadratic formula. Selecting $a=2$, $b=-5$ and $c=-6$, the solutions are given by

$\begin{array}{c}x&=&\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=&\frac{5\pm\sqrt{(-5)^2-4\times 2\times -6}}{2\times 2}\\&=&\frac{5\pm\sqrt{73}}{4}\\&\approx&3.39, -0.89\end{array}$

Taking out a factor of 2 gives the equation:

$\begin{array}{c}2\left(x^2+3x-\frac{9}{2}\right)&=&0\\\Rightarrow \hspace{5pt}2\left(\left(x+\frac{3}{2}\right)^2-\frac{27}{4}\right)&=&0\\\Rightarrow \hspace{5pt}\left(x+\frac{3}{2}\right)^2&=&\frac{27}{4}\\\Rightarrow \hspace{5pt}x&=&-\frac{3}{2}\pm\sqrt{\frac{27}{4}}\end{array}$.

Hence, $x=-\frac{3}{2}+\frac{3\sqrt{3}}{2}$ and $x= -\frac{3}{2}-\frac{3\sqrt{3}}{2}$.