To solve the equation \( 7 x^{\frac{2}{3}} - 34 x^{\frac{1}{3}} - 5 = 0 \), we can proceed as follows:
1. Substitution:
Let's set \( u = x^{\frac{1}{3}} \). Consequently, we have:
[tex]\[
u^2 = (x^{\frac{1}{3}})^2 = x^{\frac{2}{3}}
\][/tex]
Using this substitution, our equation becomes:
[tex]\[
7u^2 - 34u - 5 = 0
\][/tex]
2. Solve the Quadratic Equation:
We can solve the quadratic equation \( 7u^2 - 34u - 5 = 0 \) using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 7 \), \( b = -34 \), and \( c = -5 \).
Plugging in the values, we get:
[tex]\[
u = \frac{-(-34) \pm \sqrt{(-34)^2 - 4 \cdot 7 \cdot (-5)}}{2 \cdot 7}
\][/tex]
Simplifying further:
[tex]\[
u = \frac{34 \pm \sqrt{1156 + 140}}{14}
\][/tex]
[tex]\[
u = \frac{34 \pm \sqrt{1296}}{14}
\][/tex]
[tex]\[
u = \frac{34 \pm 36}{14}
\][/tex]
3. Find the Solutions for \( u \):
This gives us two potential solutions for \( u \):
[tex]\[
u_1 = \frac{34 + 36}{14} = \frac{70}{14} = 5
\][/tex]
[tex]\[
u_2 = \frac{34 - 36}{14} = \frac{-2}{14} = -\frac{1}{7}
\][/tex]
4. Back-Substitute \( u \) to Solve for \( x \):
Recall that \( u = x^{\frac{1}{3}} \). Thus, we need to solve for \( x \).
For \( u = 5 \):
[tex]\[
5 = x^{\frac{1}{3}}
\][/tex]
Cubing both sides:
[tex]\[
x = 5^3 = 125
\][/tex]
For \( u = -\frac{1}{7} \):
[tex]\[
-\frac{1}{7} = x^{\frac{1}{3}}
\][/tex]
Cubing both sides:
[tex]\[
x = \left(-\frac{1}{7}\right)^3 = -\frac{1}{343}
\][/tex]
Thus, the solutions for the original equation [tex]\( 7 x^{\frac{2}{3}} - 34 x^{\frac{1}{3}} - 5 = 0 \)[/tex] are [tex]\( x = 125 \)[/tex] and [tex]\( x = -\frac{1}{343} \)[/tex].
