Question

Solve each of the following equations for x, giving your answer correct to 3 decimal places:

(ii) $x^2-16x+6=0$

Answer 1

Compare $x^2-16x+6=0$ with $a{x}^2+bx+c=0$

$a=1,b=-16andc=6$

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

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

$\Rightarrow x=\frac{-(-16)\pm \sqrt{(-16{)}^2-4(1\left)\right(6)}}{2\left(1\right)}$

$\Rightarrow x=\frac{16\pm \sqrt{256-24}}{2}$

$\Rightarrow x=\frac{16\pm \sqrt{232}}{2}$

$\Rightarrow x=\frac{16\pm 15.231}{2}$

$\Rightarrow x=\frac{16+15.231}{2}andx=\frac{16-15.231}{2}$

$\therefore x=15.616and0.384$.