Question

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

Answer 1

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

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

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

$a=1,b=-3,c=-9$

We know, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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

$=\frac{3\pm \sqrt{9+36}}{2}=\frac{3\pm \sqrt{45}}{2}$

$=\frac{3\pm 3\sqrt{5}}{2}=\frac{3\pm 3\times 2.236}{2}$

$x=\frac{3\pm 6.708}{2}$

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

$x=4.85$ and $-1.85$