Chapter 7.2, Problem 32WE

To determine

Tosolve:The given radical equation.

Answer to Problem 32WE

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

  $\begin{array}{l}{u}_1=\left(-\sqrt{3}+\sqrt{6}\right)\\ u_2=\left(-\sqrt{3}-\sqrt{6}\right)\end{array}$ .

Explanation of Solution

Given:

The given radical equation is

  $u^2+2u\sqrt{3}-3=0$

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:

  $\begin{array}{l}a=1\\ b=2\sqrt{3}\\ c=-3\end{array}$

Substituting these in the formula for the roots.

First take the positive sign.

  $\begin{array}{c}{u}_1=\frac{\left(-2\sqrt{3}\right)+\sqrt{{\left(2\sqrt{3}\right)}^2-4\times 1\times \left(-3\right)}}{2\times 1}\\ =\frac{-2\sqrt{3}+\sqrt{12+12}}{2}\\ =\frac{2\left(-\sqrt{3}+\sqrt{6}\right)}{2}\\ =\left(-\sqrt{3}+\sqrt{6}\right)\end{array}$

Take the negative sign.

  $\begin{array}{c}{u}_2=\frac{\left(-2\sqrt{3}\right)-\sqrt{{\left(2\sqrt{3}\right)}^2-4\times 1\times \left(-3\right)}}{2\times 1}\\ =\frac{-2\sqrt{3}-\sqrt{12+12}}{2}\\ =\frac{2\left(-\sqrt{3}-\sqrt{6}\right)}{2}\\ =\left(-\sqrt{3}-\sqrt{6}\right)\end{array}$

Therefore, the solution of the given radical equation is

  $\begin{array}{l}{u}_1=\left(-\sqrt{3}+\sqrt{6}\right)\\ u_2=\left(-\sqrt{3}-\sqrt{6}\right)\end{array}$ .