Question

With the help of the flow chart given below solve the equation $x^2+2\sqrt{3}x+3=0$ using the formula.

Answer 1

$$\begin{gathered} \boxed{\mathrm{Compare}\mathrm{equations} \\ {x}^2+2\sqrt{3}x+3=0\mathrm{and} \\ a{x}^2+bx+c=0\mathrm{find} \\ \mathrm{the}\mathrm{values}\mathrm{of}a,b,c.}\to \boxed{\mathrm{Find}\mathrm{value}\mathrm{of} \\ {b}^2-4ac}\to \boxed{\mathrm{Write}\mathrm{formula} \\ \mathrm{to}\mathrm{solve} \\ \mathrm{quadratic} \\ \mathrm{equation}.}\to \boxed{\mathrm{Substitute} \\ \mathrm{values}\mathrm{of} \\ a,b,c\mathrm{and} \\ \mathrm{find}\mathrm{roots}} \end{gathered}$$

comparing $x^2+2\sqrt{3}x+3=0$ with $a{x}^2+bx+c=0$ we get
$a=1,b=2\sqrt{3}\mathrm{and}c=3$
$b^2-4ac={\left(2\sqrt{3}\right)}^2-4\times 1\times 3=12-12=0$
Formula to solve a quadratic equation will be
$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \Rightarrow x=\frac{-2\sqrt{3}\pm \sqrt{0}}{2\times 1}=\frac{-2\sqrt{3}}{2}=-\sqrt{3} \end{gathered}$$
Thus, $x=-\sqrt{3},-\sqrt{3}$.