Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite[latex]\,{8}^{\frac{2}{3}}\,[/latex]as[latex]\,{\left({8}^{\frac{1}{3}}\right)}^{2}.[/latex]

The way to remove the exponent on x is by raising both sides of the equation to a power that is the reciprocal of[latex]\,\frac{5}{4},[/latex]which is[latex]\,\frac{4}{5}.[/latex]

This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero.

Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the lowest exponent. Rewrite[latex]\,{x}^{\frac{1}{2}}\,[/latex]as[latex]\,{x}^{\frac{2}{4}}.\,[/latex]Then, factor out[latex]\,{x}^{\frac{2}{4}}\,[/latex]from both terms on the left.

Where did[latex]\,{x}^{\frac{1}{4}}\,[/latex]come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply[latex]\,{x}^{\frac{2}{4}}\,[/latex]back in using the distributive property, we get the expression we had before the factoring, which is what should happen. We need an exponent such that when added to[latex]\,\frac{2}{4}\,[/latex]equals[latex]\,\frac{3}{4}.\,[/latex]Thus, the exponent on x in the parentheses is[latex]\,\frac{1}{4}.\,[/latex]

Let us continue. Now we have two factors and can use the zero factor theorem.

The two solutions are [latex]\,0[/latex] and [latex]\frac{1}{81}.[/latex]

First, set the equation equal to zero. Then factor out what is common to both terms, the GCF.

Notice that we have the difference of squares in the factor[latex]\,{x}^{2}-16,[/latex]which we will continue to factor and obtain two solutions. The first term,[latex]\,5{x}^{2},[/latex]generates, technically, two solutions as the exponent is 2, but they are the same solution.

The solutions are [latex]\,0\text{ (double solution),}[/latex][latex]4,[/latex] and [latex]\,-4.[/latex]

This polynomial consists of 4 terms, which we can solve by grouping. Grouping procedures require factoring the first two terms and then factoring the last two terms. If the factors in the parentheses are identical, we can continue the process and solve, unless more factoring is suggested.

The grouping process ends here, as we can factor[latex]\,{x}^{2}-9\,[/latex]
using the difference of squares formula.

Figure 2.

The solutions are [latex]3,[/latex][latex]-3,[/latex] and [latex]\,-1.\,[/latex]Note that the highest exponent is 3 and we obtained 3 solutions. We can see the solutions, the x-intercepts, on the graph in (Figure).

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 [latex]-5\,[/latex] and [latex]3.\,[/latex]Let us check each solution back in the original equation. First, check[latex]\,x=-5.[/latex]

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[latex]\,x=3.[/latex]

The solution is [latex]\,3.[/latex]

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:[latex]\,{\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}.[/latex]

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

The proposed solutions are [latex]\,3\,[/latex] and [latex]\,83.\,[/latex]Check each solution in the original equation.

One solution is [latex]\,3.[/latex]

Check[latex]\,x=83.[/latex]

The only solution is [latex]\,3.\,[/latex]We see that[latex]\,x=83\,[/latex]is an extraneous solution.

• (a) [latex]|6x+4|=8[/latex]

  Write two equations and solve each:

  [latex]\begin{array}{ccccccc}\hfill 6x+4& =& 8\hfill & \phantom{\rule{5em}{0ex}}& \hfill 6x+4& =& -8\hfill \\ \hfill 6x& =& 4\hfill & \phantom{\rule{5em}{0ex}}& \hfill 6x& =& -12\hfill \\ \hfill x& =& \frac{2}{3}\hfill & \phantom{\rule{5em}{0ex}}& \hfill x& =& -2\hfill \end{array}[/latex]

  The two solutions are [latex]\,\frac{2}{3}[/latex] and [latex]-2.[/latex]

• (b) [latex]|3x+4|=-9[/latex]

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

• (c) [latex]|3x-5|-4=6[/latex]

  Isolate the absolute value expression and then write two equations.

  [latex]\begin{array}{ccccccccc}& & & \hfill |3x-5|-4& =& 6\hfill & & & \\ & & & \hfill |3x-5|& =& 10\hfill & & & \\ \hfill 3x-5& =& 10\hfill & & & & \hfill 3x-5& =& -10\hfill \\ \hfill 3x& =& 15\hfill & & & & \hfill 3x& =& -5\hfill \\ \hfill x& =& 5\hfill & & & & \hfill x& =& -\frac{5}{3}\hfill \end{array}[/latex]

  There are two solutions: [latex]\,5,[/latex] and [latex]-\frac{5}{3}.[/latex]

• (d) [latex]|-5x+10|=0[/latex]

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

  [latex]\begin{array}{ccc}\hfill -5x+10& =& 0\hfill \\ \hfill -5x& =& -10\hfill \\ \hfill x& =& 2\hfill \end{array}[/latex]

  There is one solution: [latex]\,2.[/latex]