Question

How to solve for x if $ {{x}^{\dfrac{2}{3}}}-5{{x}^{\dfrac{1}{3}}}+6=0 $ ?

Answer 1

Hint: In such questions, students usually get confused about how to start the problem. This question can be solved using a substitution method. We can substitute u for $ {{x}^{\dfrac{1}{3}}} $ and then solve the equation as a quadratic equation using the method of splitting the middle term.

Complete step by step answer:
According to the question, the given equation is $ {{x}^{\dfrac{2}{3}}}-5{{x}^{\dfrac{1}{3}}}+6=0 $ and we have been asked to find the value of x.
Let us consider u = $ {{x}^{\dfrac{1}{3}}} $
Therefore, we get the equation as
$\Rightarrow$ $ {{u}^{2}}-u-6=0 $
Now we will solve this quadratic equation using splitting the middle term method. In this method, we split the middle term which is the sum of two factors and the product of these two factors is equal to the product of the first term and last term of the quadratic equation.
Therefore, we get
$\Rightarrow$ $ {{u}^{2}}-u-6=0 $
Now we will split the middle term u as -3u + 2u. 3u and 2u is also equal to the product of the first and last term $ 6{{u}^{2}} $
Therefore, we get
 $ \Rightarrow {{u}^{2}}-3u+2u-6=0 $
Now we will take u common from the above equation, therefore, we get
$ \Rightarrow u(u-3)+2(u-3)=0 $
Now taking (u – 3) common from the above equation, we get
$\Rightarrow$ $ \Rightarrow (u-3)(u+2)=0 $
Now again substituting $ {{x}^{\dfrac{1}{3}}} $ = u , we get
$\Rightarrow$ $ ({{x}^{\dfrac{1}{3}}}-3)({{x}^{\dfrac{1}{3}}}+2)=0 $
Now we know that $ ({{x}^{\dfrac{1}{3}}}-3)=0 $ and $ ({{x}^{\dfrac{1}{3}}}+2)=0 $
Therefore taking $ ({{x}^{\dfrac{1}{3}}}-3)=0 $ first, we get
$\Rightarrow$ $ {{x}^{\dfrac{1}{3}}}=3 $
Cubing both sides, we get
 $ \begin{align}
  & {{({{x}^{\dfrac{1}{3}}})}^{3}}={{(3)}^{3}} \\
 & \Rightarrow x=27 \\
\end{align} $
Now taking $ ({{x}^{\dfrac{1}{3}}}+2)=0 $ , we get
 $ {{x}^{\dfrac{1}{3}}}=-2 $
Cubing both sides, we get
 $ \begin{align}
  & {{({{x}^{\dfrac{1}{3}}})}^{3}}={{(-2)}^{3}} \\
 & \Rightarrow x=-8 \\
\end{align} $
Therefore, we get the values of x for the equation as x = 27 or x = -8.

Note:
 We can also solve this question by using the discriminant formula of quadratic equations after substituting u for $ {{x}^{\dfrac{1}{3}}} $ otherwise we cannot solve this question. Also, while solving the question we need to be very careful because in a hurry we might end up reading the given equation wrong and end up getting the wrong answer.