Chapter 1, Problem 40REAE

Solution

Solve the quadratic equation $6x^2-8x=-3$ for $\left(x\right)$ by quadratic formula.

The solution for the quadratic equation is $x=\left\{\frac{4+i\sqrt{2}}{6},\frac{4-i\sqrt{2}}{6}\right\}$ .

Given:

The quadratic equation is $6x^2-8x=-3$ .

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 quadratic equation $6x^2-8x=-3$ is re-written as:

  $\begin{array}{c}6x^2-8x=-3\mathrm{......}\left(\text{Given}\right)\\ 6x^2-8x+3=0\mathrm{......}\left(2\right)\end{array}$

The solution $\left(x\right)$ for the quadratic equation $6x^2-8x+3=0$ from $\left(2\right)$ can be evaluated with the help of formula $\left(1\right)$ . Therefore from $\left(1\right)$ ,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\left(6\right)\left(3\right)}}{2\left(6\right)}\mathrm{......}\left(\because 6x^2-8x+3=0\right)\\ =\frac{8\pm i2\sqrt{2}}{12}\\ =\frac{8+i2\sqrt{2}}{12},\frac{8-i2\sqrt{2}}{12}\\ =\left\{\frac{4+i\sqrt{2}}{6},\frac{4-i\sqrt{2}}{6}\right\}\end{array}$

Hence the solution for the quadratic equation is $x=\left\{\frac{4+i\sqrt{2}}{6},\frac{4-i\sqrt{2}}{6}\right\}$ .