Chapter 1.8, Problem 9E

Solution

Solve the quadratic equation $4x^2-8x+1=0$ for $\left(x\right)$ .

The solution for the quadratic equation is $x=\left\{\frac{2+\sqrt{3}}{2},\frac{2-\sqrt{3}}{2}\right\}$ .

Given:

The quadratic equation is $4x^2-8x+1=0$ .

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 $4x^2-8x+1=0$ can be evaluated with the help of formula $\left(1\right)$ . Therefore,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\left(4\right)\left(1\right)}}{2\left(4\right)}\mathrm{......}\left(\because 4x^2-8x+1=0\right)\\ =\frac{8\pm \sqrt{48}}{8}\\ =\frac{8\pm 4\sqrt{3}}{8}\\ =\frac{8+4\sqrt{3}}{8},\frac{8-4\sqrt{3}}{8}\\ =\left\{\frac{2+\sqrt{3}}{2},\frac{2-\sqrt{3}}{2}\right\}\end{array}$

Hence the solution for the quadratic equation is $x=\left\{\frac{2+\sqrt{3}}{2},\frac{2-\sqrt{3}}{2}\right\}$ .