Question

Solve the given quadratic equation, $x^2-3(x+3)=0$; giving your answer correct to two significant figures.

Answer 1

Given equation: $x^2-3(x+3)=0$

$\Rightarrow x^2-3x-9=0$

Comparing with $a{x}^2+bx+c=0$ we get, $a=1,b=-3$ and $c=-9$

We know that, for the equation $a{x}^2+bx+c=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ are roots of the equation.

$\Rightarrow x=\frac{-(-3)\pm \sqrt{(-3{)}^2-4(1\left)\right(-9)}}{2\times 1}$

$\Rightarrow x=\frac{3\pm \sqrt{9+36}}{2}$

$\Rightarrow x=\frac{3\pm \sqrt{45}}{2}=\frac{3\pm 6.708}{2}$

$\Rightarrow x=\frac{3+6.708}{2}$ or $x=\frac{3-6.708}{2}$

$\Rightarrow x=\frac{9.708}{2}$ or $x=\frac{-3.708}{2}$

$\Rightarrow x=4.854$ or $x=-1.854$

$\therefore x=4.9$ or $x=-1.9$ corrected to two significant figures.