Question

How do you solve ${x^{\dfrac{2}{3}}} - 3{x^{\dfrac{1}{3}}} - 4 = 0$?

Answer 1

Hint: We will first assume that $y = {x^{\dfrac{1}{3}}}$ and thus, we will obtain a quadratic equation in y. Now, we will use the quadratic formula and thus, we will have the required answer by putting back the assumed $y = {x^{\dfrac{1}{3}}}$.

Complete step by step solution:
We are given that we are required to solve ${x^{\dfrac{2}{3}}} - 3{x^{\dfrac{1}{3}}} - 4 = 0$. …………….(1)
Let us assume that $y = {x^{\dfrac{1}{3}}}$ ………….(2)
Squaring both sides of the equation number 2, we will then obtain the following equation:-
$ \Rightarrow {y^2} = {x^{\dfrac{2}{3}}}$ …………….(3)
Putting the equation number 2 and 3 in equation number 1, we will then obtain the following equation:-
$ \Rightarrow {y^2} - 3y - 4 = 0$
We know that the general quadratic equation is given by the equation $a{x^2} + bx + c = 0$, where a, b and c are constants.
The roots of this general equation is given by the following expression:-
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing the equation to the general equation, we have: a = 1, b = - 3 and b = - 4.
Therefore, the roots are: $y = \dfrac{{ - ( - 3) \pm \sqrt {{{( - 3)}^2} - 4(1)( - 4)} }}{{2(1)}}$
Simplifying the calculations in the right side of the above equation, we will then obtain the following equation:-
$ \Rightarrow y = \dfrac{{3 \pm \sqrt {9 + 16} }}{2}$
Simplifying the calculations in the right side of the above equation further, we will then obtain the following equation:-
$ \Rightarrow y = \dfrac{{3 \pm \sqrt {25} }}{2}$
Since we know that the square – root of 25 is 5, therefore, we have the following equation:-
$ \Rightarrow y = \dfrac{{3 \pm 5}}{2}$
Therefore, the roots are y = 4, - 1.
Now, since we assumed that $y = {x^{\dfrac{1}{3}}}$.
$ \Rightarrow {x^{\dfrac{1}{3}}} = - 1,4$
Taking cubes on both the sides, we will then obtain the following equation:-
 $ \Rightarrow x = - 1,64$
Thus, we have the required answer.

Note: The students must note that there is an alternate way to solve the question as well.
Alternate Way:
We are given that we are required to solve ${x^{\dfrac{2}{3}}} - 3{x^{\dfrac{1}{3}}} - 4 = 0$.
Let us assume that $y = {x^{\dfrac{1}{3}}}$
Therefore, we obtain the following equation:-
$ \Rightarrow {y^2} - 3y - 4 = 0$
Now, we will use the method of ‘splitting the middle term’. So, we can write it as follows:-
$ \Rightarrow {y^2} + y - 4y - 4 = 0$
Taking y common from first two terms and – 4 common from last two terms, we will then obtain the following equation:-
$ \Rightarrow y(y + 1) - 4(y + 1) = 0$
Taking (y + 1) common from both the terms above, we will obtain:-
$ \Rightarrow (y + 1)(y - 4) = 0$
Therefore, the roots are y = 4, - 1.
Now, since we assumed that $y = {x^{\dfrac{1}{3}}}$.
$ \Rightarrow {x^{\dfrac{1}{3}}} = - 1,4$
Taking cubes on both the sides, we will then obtain the following equation:-
 $ \Rightarrow x = - 1,64$