Question

Find the value of \[x\] if the given expression is \[{x^{\dfrac{2}{3}}} - 7{x^{\dfrac{1}{3}}} + 10 = 0\]
A. \[125\]
B. \[8\]
c. \[\phi \]
D. \[125,8\]

Answer 1

Hint: The given equation is an algebraic equation. The left-hand side is an algebraic equation in terms of the variable \[x\], and the right-hand side is zero, and we have to find the value of \[x\]. So, we have to find the value of \[x\] at which the value of the left-hand side of the algebraic equation comes out to be zero. That means we have to find the solution to the given equation. But the exponents of the variable are fractional, which makes it difficult to find the value of \[x\] so we will have to use substitution to make this equation a quadratic equation. Then we find the solution of the quadratic equation by using factorization.

Formula used: The following formula will be useful for this question
\[\begin{array}{l}{x^a} = y\\ \Rightarrow x = {y^{\dfrac{1}{a}}}\end{array}\]

Complete step-by-step solution:
We are given an algebraic equation \[{x^{\dfrac{2}{3}}} - 7{x^{\dfrac{1}{3}}} + 10 = 0\]
Firstly, we will simplify the equation
Let \[{x^{\dfrac{1}{3}}} = y\] , so \[{x^{\dfrac{2}{3}}} = {y^2}\]
Thus the equation becomes
\[{y^2} - 7y + 10 = 0\]
Now, we will solve this quadratic equation by using the factoring method:
We have to split the middle term \[7y\] such that the sum of the coefficients of \[y\] is equal to \[7\] and their product is equal to \[10\].
\[\begin{array}{l}{y^2} - 2y - 5y + 10 = 0\\ y(y - 2) - 5(y - 2) = 0\\(y - 2)(y - 5) = 0\\y = 2,5\end{array}\]
Now, we will put the value of \[y\] in the \[{x^{\dfrac{1}{3}}} = y\]
Firstly we will solve for\[y = 2\]
\[\begin{array}{l}{x^{\dfrac{1}{3}}} = 2\\x = {(2)^3}\\x = 8\end{array}\]
Now, we will solve for \[y = 5\]
\[\begin{array}{l}\,{x^{\dfrac{1}{3}}} = 5\\x = {(5)^3}\\x = 125\end{array}\]
Hence the value of the expression is \[125, 8\]

Therefore, the option (D) is correct.

Note: In the above solution, we have used factorization to find the solution of the quadratic equation. There are many other methods that we can use to find the solution to a quadratic equation such as completing the square method, square roots, and the quadratic formula. We choose the most convenient method to solve a quadratic equation.