Answer :
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].
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].
