2.6 Other types of equations  (Page 3/10)

The radical is already isolated on the left side of the equal side, so proceed to square both sides.

We see that the remaining equation is a quadratic. Set it equal to zero and solve.

The proposed solutions are $\mathrm{-5}$ and $3.$ Let us check each solution back in the original equation. First, check $x=\mathrm{-5.}$

This is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation.

Check $x=3.$

The solution is $3.$

As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.

Use the perfect square formula to expand the right side: ${\left(a-b\right)}^2=a^2\mathrm{-2}ab+b^2.$

Now that both radicals have been eliminated, set the quadratic equal to zero and solve.

The proposed solutions are $3$ and $83.$ Check each solution in the original equation.

One solution is $3.$

Check $x=83.$

The only solution is $3.$ We see that $x=83$ is an extraneous solution.

• (a) $\left|6x+4\right|=8$

  Write two equations and solve each:

  $\begin{array}{ccccccc}\hfill 6x+4& =& 8\hfill & & \hfill 6x+4& =& \mathrm{-8}\hfill \\ \hfill 6x& =& 4\hfill & & \hfill 6x& =& \mathrm{-12}\hfill \\ \hfill x& =& \frac{2}{3}\hfill & & \hfill x& =& \mathrm{-2}\hfill \end{array}$

  The two solutions are $\frac{2}{3}$ and $\mathrm{-2.}$

• (b) $\left|3x+4\right|=\mathrm{-9}$

  There is no solution as an absolute value cannot be negative.

• (c) $\left|3x-5\right|-4=6$

  Isolate the absolute value expression and then write two equations.

  $$\begin{array}{ccccccccc}& & & \hfill \left|3x-5\right|-4& =& 6\hfill & & & \\ & & & \hfill \left|3x-5\right|& =& 10\hfill & & & \\ \hfill 3x-5& =& 10\hfill & & & & \hfill 3x-5& =& \mathrm{-10}\hfill \\ \hfill 3x& =& 15\hfill & & & & \hfill 3x& =& \mathrm{-5}\hfill \\ \hfill x& =& 5\hfill & & & & \hfill x& =& -\frac{5}{3}\hfill \end{array}$$

  There are two solutions: $5,$ and $-\frac{5}{3}.$

• (d) $\left|\mathrm{-5}x+10\right|=0$

  The equation is set equal to zero, so we have to write only one equation.

  $$\begin{array}{ccc}\hfill \mathrm{-5}x+10& =& 0\hfill \\ \hfill \mathrm{-5}x& =& \mathrm{-10}\hfill \\ \hfill x& =& 2\hfill \end{array}$$

  There is one solution: $2.$