Chapter 1.8, Problem 1GP

Solution

Solve the quadratic equation $x^2=6x-4$ for $\left(x\right)$ .

The solution for the quadratic equation is $x=\left\{3+\sqrt{5},3-\sqrt{5}\right\}$ .

Given:

The quadratic equation is $x^2=6x-4$ which can be written as $x^2-6x+4=0$ .

Concept used: Any quadratic equation of the form $a{x}^2+bx+c=0\text{ where }a\ne 0$ has the roots $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(1\right)$

Calculation:

The value of $\left(x\right)$ for the quadratic equation $x^2-6x+4=0$ can be evaluated with the help of formula $\left(1\right)$ . Therefore,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(1\right)\left(4\right)}}{2\left(1\right)}\mathrm{......}\left(\because x^2-6x+4=0\right)\\ =\frac{6\pm \sqrt{20}}{2}\\ =\frac{6\pm 2\sqrt{5}}{2}\\ =3\pm \sqrt{5}\\ =\left\{3+\sqrt{5},3-\sqrt{5}\right\}\end{array}$

Hence the solution for the quadratic equation is $x=\left\{3+\sqrt{5},3-\sqrt{5}\right\}$ .