Question

How do you solve $ 3{x^{\dfrac{2}{3}}} + {x^{\dfrac{1}{3}}} - 2 = 0 $ ?

Answer 1

Hint: The given question can be solved by first substituting the given variable which has its power given in fraction to a variable that will become quadratic since the first power is two times the power on the next term. Also then we will solve the quadratic equation for that variable using either factorization method(if the quadratic equation is easily factorizable ) or we will use the quadratic equation formula which will then give us the value of the two roots the given roots will then be easily transformed back into the original variable (in the variable the question has been asked). And we will get the value of our two roots in the original variable.

Complete step-by-step answer:
Given, $ 3{x^{\dfrac{2}{3}}} + {x^{\dfrac{1}{3}}} - 2 = 0 $
We will now get rid of these fractional powers by method of substitution, we substitute,
 $ w = {x^{\dfrac{1}{3}}} $
Now we will transform the above equation into the new variable $ w $ .
The question will then become
 $ 3{\left( w \right)^2} + 1 \cdot {\left( w \right)^1} - 2 = 0 $
The above equation is a quadratic equation is $ w $ we will now solve the given equation using the quadratic formula,
 $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ where $ a,b,c $ are the coefficients of the quadratic equation at our hand.
The equation will be solved as
 $ w = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4*3*\left( { - 2} \right)} }}{{2*3}} $
The value of $ w $ from solving this equation is given as
 $ w = \dfrac{{ - 1 \pm 5}}{6} $
Thus the two values of $ w $ are as follows:
 $ w = - 1,\dfrac{2}{3} $
The given roots of the equations will now be transformed back into the original variable which is given as
 $ x = \dfrac{8}{{27}}, - 1 $
Thus the above are the roots of the given equation $ 3{x^{\dfrac{2}{3}}} + {x^{\dfrac{1}{3}}} - 2 = 0 $
So, the correct answer is “ $ x = \dfrac{8}{{27}}, - 1 $”.

Note: When solving questions like this of changing one variable into another always remember to change back the roots into the original variable’s roots to reduce unfortunate deduction of marks, this seems simple enough but is forgotten in the crunch moments so the student should make it a habit.