1.6 Other Types of Equations, Equations, Equations that are Quadratic in Form

Quadratic in Form (U-substitution)

Example: Solve the equation.

5x^{2/3}-6x^{1/3}+1=0

Solution:

The equation is similar to a quadratic.  It has 3 terms and one exponent is twice the other.  Since the equation is quadratic in form, use substitution to solve the equation.

Use the following substitution to rewrite the equation

u=x^{1/3}

u^2=x^{2/3}

Original Equation
5x^{2/3}-6x^{1/3}+1=0
Substitute
u=x^{1/3}
u^2=x^{2/3}
5x^{2/3}-6x^{1/3}+1=0
5u^{2}-6u+1=0
Solve the quadratic equation by factoring.
1) Factor the quadratic
5u^{2}-6u+1=0
(5u-1)(u-1)=0
Solve the quadratic equation by factoring.
2) Apply the zero product property
(5u-1)(u-1)=0
5u-1=0 or u-1=0
Solve the quadratic equation by factoring.
3) Solve each linear factor
5u-1=0 or u-1=0
5u-1+1=0+1 or u-1+1=0+1
5u=1 or u=1
{5u}/5=1/5 or u=1
u=1/5 or u=1
Substitute again to bring back the original variable.  Use the original substitution.
u=x^{1/3}
u=1/5 or u=1
x^{1/3}=1/5 or x^{1/3}=1 
Solve the equation with rational exponents.
1) Rewrite the rational exponents in radical form
x^{1/3}=1/5 or x^{1/3}=1
root{3}{x}=1/5 or root{3}{x}=1
Solve the equation with rational exponents.
2) Cancel the cube root by cubing both sides.
3) Simplify
root{3}{x}=1/5 or root{3}{x}=1
(root{3}{x})^3=(1/5)^3 or (root{3}{x})^3=(1)^3
x=1/125 or x=1

The solution to 5x^{2/3}-6x^{1/3}+1=0 isx=1/125 or x=1.

 

Here is a video with similar examples.