Question

The solution set of the equation ${x^{\dfrac{2}{3}}} + {x^{\dfrac{1}{3}}} = 2$ is
A. {-8, 1}
B. {8, 1}
C. {1, -1}
D. {2, -2}

Answer 1

Hint: We cannot directly solve the given equation. So we will substitute ${x^{\dfrac{1}{3}}} = y$ and get a new equation. By substituting these values, we will get a quadratic equation in terms of y. Now we can solve this quadratic equation using the quadratic equation. After getting the values of y we can find the values of x easily.

Complete step-by-step solution:
In this question, we are given an equation with power in fractions and we are supposed to find its solution set.
Given equation: ${x^{\dfrac{2}{3}}} + {x^{\dfrac{1}{3}}} = 2$- - - - - - - - - - (1)
Now, we cannot solve this equation directly. So, we will be substituting ${x^{\dfrac{1}{3}}} = y$- - - - - - - (2)
Now, we can write ${x^{\dfrac{2}{3}}} = {x^{2 \times \dfrac{1}{3}}} = {\left( {{x^{\dfrac{1}{3}}}} \right)^2}$. Now, we can substitute ${x^{\dfrac{1}{3}}} = y$ in the first term too.
Therefore, equation (1) becomes
$ \Rightarrow {y^2} + y = 2$
Taking 2 from RHS to LHS, we get
$ \Rightarrow {y^2} + y - 2 = 0$- - - - - - - - - (3)
Now, equation (3) is a quadratic equation with variable y. So, we can solve this equation using the quadratic formula. The quadratic formula is
 $ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here, $a = 1,b = 1,b = - 2$. Therefore,
$ \Rightarrow y = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4\left( 1 \right)\left( { - 2} \right)} }}{{2\left( 1 \right)}}$
$
   \Rightarrow y = \dfrac{{ - 1 \pm \sqrt {1 + 8} }}{2} \\
   \Rightarrow y = \dfrac{{ - 1 \pm \sqrt 9 }}{2} \\
   \Rightarrow y = \dfrac{{ - 1 \pm 3}}{2} \\
 $
$
   \Rightarrow y = \dfrac{{ - 1 + 3}}{2} \\
   \Rightarrow y = \dfrac{2}{2} \\
   \Rightarrow y = 1 \\
 $ and $
   \Rightarrow y = \dfrac{{ - 1 - 3}}{2} \\
   \Rightarrow y = \dfrac{{ - 4}}{2} \\
   \Rightarrow y = - 2 \\
 $
Therefore, $y = 1, - 2$.
Now, we need to find the value of x.
Therefore, from equation (2), we get
$ \Rightarrow {x^{\dfrac{1}{3}}} = y$
Now, for $y = 1$, we get
$
   \Rightarrow {x^{\dfrac{1}{3}}} = 1 \\
   \Rightarrow x = 1 \\
 $
And for $y = - 2$, we get
$ \Rightarrow {x^{\dfrac{1}{3}}} = - 2$
Now, multiplying the powers on both LHS and RHS with 3, we get
\[
   \Rightarrow {x^{\dfrac{1}{3} \times 3}} = - {2^3} \\
   \Rightarrow x = - 8 \\
 \]
Hence, the solution set of ${x^{\dfrac{2}{3}}} + {x^{\dfrac{1}{3}}} = 2$ is $\left\{ { - 8,1} \right\}$.
Hence, the correct answer is option A.

Note: We can cross verify our answer by putting the values of x in the given equation.
$ \Rightarrow {x^{\dfrac{2}{3}}} + {x^{\dfrac{1}{3}}} = {\left( { - 8} \right)^{\dfrac{2}{3}}} + {\left( { - 8} \right)^{\dfrac{1}{3}}}$
                    $
   = {\left\{ {{{\left( { - 8} \right)}^{\dfrac{1}{3}}}} \right\}^2} + {\left( { - 8} \right)^{\dfrac{1}{3}}} \\
   = {\left( { - 2} \right)^2} + \left( { - 2} \right) \\
   = 4 - 2 \\
   = 2 \\
 $
$ \Rightarrow {x^{\dfrac{2}{3}}} + {x^{\dfrac{1}{3}}} = {\left( 1 \right)^{\dfrac{2}{3}}} + {\left( 1 \right)^{\dfrac{1}{3}}}$
                     $
   = 1 + 1 \\
   = 2 \\
 $
Hence, both our values are correct.