Expand menu menu_open Minimize Start chapters Home History history History expand_more
{{ item.displayTitle }}
navigate_next
No history yet!
Progress & Statistics equalizer Progress expand_more
Student
navigate_next
Teacher
navigate_next
{{ filterOption.label }}
{{ item.displayTitle }}
{{ item.subject.displayTitle }}
arrow_forward
No results
{{ searchError }}
search
menu_open
{{ courseTrack.displayTitle }}
{{ statistics.percent }}% Sign in to view progress
{{ printedBook.courseTrack.name }} {{ printedBook.name }}
search Use offline Tools apps
Login account_circle menu_open

Solving Radical Equations

Solving Radical Equations 1.6 - Solution

arrow_back Return to Solving Radical Equations

Solving a radical equation usually involves three main steps.

  1. Isolate the radical on one side of the equation.
  2. Raise each side of the equation to a power equal to the index of the radical to eliminate the radical.
  3. Solve the resulting equation.
  4. Check the results for extraneous solutions.
Now we can analyze the given radical equation. 810x315=17\begin{gathered} 8\sqrt[3]{10x}-15=17 \end{gathered} First, let's isolate the radical, 10x3,\sqrt[3]{10x}, on one side of the equation.
810x315=178\sqrt[3]{10x}-15=17
810x3=328\sqrt[3]{10x}=32
10x3=4\sqrt[3]{10x}=4
We got an isolated radical with index equal to 3.{\color{#0000FF}{3}}. Then, we will raise each side of the equation to the power of 3.{\color{#0000FF}{3}}.
10x3=4\sqrt[3]{10x}=4
(10x3)3=43\left(\sqrt[3]{10x}\right)^3=4^3
Solve for xx
10x=4310x=4^3
10x=6410x=64
x=6410x=\dfrac{64}{10}
x=325x=\dfrac{32}{5}
Next, we will check for extraneous solutions. We do that by substituting 325\frac{32}{5} for xx in the original equation. If the substitution produces a true statement, we know that our answer is correct. If it does not, then it is an extraneous solution.
810x315=178\sqrt[3]{10x}-15=17
810(325)315=?178\sqrt[3]{10\left({\color{#0000FF}{\dfrac{32}{5}}}\right)}-15\stackrel{?}{=}17
Simplify
83205315=?178\sqrt[3]{\dfrac{320}{5}}-15\stackrel{?}{=}17
864315=?178\sqrt[3]{64}-15\stackrel{?}{=}17
8(4)15=?178(4)-15\stackrel{?}{=}17
3215=?1732-15\stackrel{?}{=}17
17=17 17=17\ {\color{#009600}{\Large\checkmark}}
Because our substitution produced a true statement, we know that our answer, x=325,x=\frac{32}{5}, is correct.