Question

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

Answer 1

Step-1: Comparing ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{)}\mathbf{=}\mathbf{0}$ with $\mathbf{}\mathit{a}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{b}\mathit{x}\mathbf{+}\mathit{c}\mathbf{=}\mathbf{0}$,

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

Step-2: Find $\mathit{x}$ using quadratic formula,

⇒ $x^2-3(x+3)=0$

⇒ $x^2-3x-9=0$

⇒ $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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

⇒ $x=\frac{3\pm \sqrt{9+36}}{2}$

⇒ $x=\frac{3\pm 6.70}{2}$

⇒ $x=\frac{9.70}{2},\frac{-3.70}{2}$

$x=4.85,-1.85$

Hence, the required solution for x are $\mathbf{4}\mathbf{.}\mathbf{85}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{85}$.