Chapter 7.2, Problem 33WE

To determine

Tosolve:The given radical equation.

Answer to Problem 33WE

The solution of the given radical equation in the simplest radical form is

  $\begin{array}{l}{x}_1=\frac{\left(-\sqrt{2}\right)}{2}\\ x_2=\left(-2\sqrt{2}\right)\end{array}$ .

Explanation of Solution

Given:

The given radical equation is

  $\sqrt{2}{x}^2+5x+2\sqrt{2}=0$

Calculation:

The solution of the given radical equation is obtained by using the quadratic formula.

For a quadratic equation in the form of $a{x}^2+bx+c=0$ , the solution of this equation is given by

  $x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Comparing the given equation with the standard quadratic equation, the coefficients are:

  $\begin{array}{l}a=\sqrt{2}\\ b=5\\ c=2\sqrt{2}\end{array}$

Substituting these in the formula for the roots.

First take the positive sign.

  $\begin{array}{c}{x}_1=\frac{\left(-5\right)+\sqrt{5^2-4\times \sqrt{2}\times \left(2\sqrt{2}\right)}}{2\times \sqrt{2}}\\ =\frac{-5+\sqrt{25-16}}{2\sqrt{2}}\\ =\frac{-5+3}{2\sqrt{2}}\\ =\frac{-2}{2\sqrt{2}}\\ =\frac{-1}{\sqrt{2}}\\ \text{Rationalizing the above expression}\\ =\frac{-1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ =-\frac{\sqrt{2}}{2}\end{array}$

Take the negative sign.

  $\begin{array}{c}{x}_2=\frac{\left(-5\right)-\sqrt{5^2-4\times \sqrt{2}\times \left(2\sqrt{2}\right)}}{2\times \sqrt{2}}\\ =\frac{-5-\sqrt{25-16}}{2\sqrt{2}}\\ =\frac{-5-3}{2\sqrt{2}}\\ =\frac{-8}{2\sqrt{2}}\\ =\frac{-4}{\sqrt{2}}\\ \text{Rationalizing the above expression}\\ =\frac{-4}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ =-2\sqrt{2}\end{array}$

Therefore, the solution of the given radical equation is

  $\begin{array}{l}{x}_1=\frac{\left(-\sqrt{2}\right)}{2}\\ x_2=\left(-2\sqrt{2}\right)\end{array}$ .