Chapter 1.8, Problem 50E

Solution

The error in solving the equation given.

The solution of the quadratic equation is $x=\left\{-3+\sqrt{3},-3-\sqrt{3}\right\}$ and the error was to get the equation first in the form of $a{x}^2+bx+c=0$ where $a\ne 0$ in first step.

Given:

The solved equation is shown below.

  $\begin{array}{c}{x}^2+6x+8=2\\ x=\frac{-6\pm \sqrt{6^2-4\left(1\right)\left(8\right)}}{2\left(1\right)}\\ =\frac{-6\pm \sqrt{4}}{2}\\ =\frac{-6\pm 2}{2}\\ =-2\text{or}-4\end{array}$

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 equation $x^2+6x+8=2$ can be re-written as $x^2+6x+6=0$ .

Now, the solution for the quadratic equation $x^2+6x+6=0$ is obtained as:

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

Hence, the solution of the quadratic equation is $x=\left\{-3+\sqrt{3},-3-\sqrt{3}\right\}$ and the error was to get the equation first in the form of $a{x}^2+bx+c=0$ where $a\ne 0$ in first step.