Question

How do you solve $6\sqrt {\dfrac{{5k}}{4}} - 3 = 0$ and check your solution?

Answer 1

Hint: In this question, we need to solve the given radical equation. Firstly, we will move all the constant terms to the right hand side by some mathematical operations. After that we take squares on both sides to eliminate the radical on the left hand side and simplify to get the desired solution. Then the obtained solution, we will substitute back in the given equation to check whether the solution is correct or not.

Complete step-by-step solution:
Given an equation of the form $6\sqrt {\dfrac{{5k}}{4}} - 3 = 0$ …… (1)
We are asked to find the solution of this equation above.
Note that the equation in (1) is one of the examples for radical equations.
A radical equation is any equation that involves any type of radical such as square root, cube root, fourth root etc.
In the beginning we try to isolate the square root on the L.H.S. in the equation (1).
Consider the given equation, $6\sqrt {\dfrac{{5k}}{4}} - 3 = 0$
Transfer -3 to the R.H.S. we get,
$ \Rightarrow 6\sqrt {\dfrac{{5k}}{4}} = 3$
Now divide throughout by 6, we get,
$ \Rightarrow \dfrac{6}{6}\sqrt {\dfrac{{5k}}{4}} = \dfrac{3}{6}$
$ \Rightarrow \sqrt {\dfrac{{5k}}{4}} = \dfrac{1}{2}$
Now we eliminate the radical on the L.H.S.
To do this, we square on both sides, we get,
$ \Rightarrow {\left( {\sqrt {\dfrac{{5k}}{4}} } \right)^2} = {\left( {\dfrac{1}{2}} \right)^2}$
Simplifying we get,
$ \Rightarrow \dfrac{{5k}}{4} = \dfrac{{{1^2}}}{{{2^2}}}$
$ \Rightarrow \dfrac{{5k}}{4} = \dfrac{1}{4}$
Since 4 is present in the numerators of both sides. Cancelling it we get,
$ \Rightarrow 5k = 1$
Dividing by 5 on both sides, we get,
$ \Rightarrow \dfrac{{5k}}{5} = \dfrac{1}{5}$
$ \Rightarrow k = \dfrac{1}{5}$.
Now we check that the solution obtained is correct by substituting back in the given equation.
We substitute $k = \dfrac{1}{5}$ in the equation (1), we get,
$ \Rightarrow 6\sqrt {\dfrac{{5 \times \dfrac{1}{5}}}{4}} - 3 = 0$
Simplifying we get,
$ \Rightarrow 6\sqrt {\dfrac{1}{4}} - 3 = 0$
This can be written as,
$ \Rightarrow 6\dfrac{{\sqrt 1 }}{{\sqrt 4 }} - 3 = 0$
We know that $\sqrt 1 = 1$ and $\sqrt 4 = 2$.
Hence we get,
$ \Rightarrow 6 \times \dfrac{1}{2} - 3 = 0$
$ \Rightarrow 3 - 3 = 0$
Therefore, our solution is correct.
Hence, the required solution for the equation $6\sqrt {\dfrac{{5k}}{4}} - 3 = 0$ is given by $k = \dfrac{1}{5}$.

Note: Students must know the meaning of the radical equation. A radical equation is any equation that involves any type of radical such as square root, cube root, fourth root etc.
Examples of radicals are : $\sqrt {35} $ and $\sqrt[3]{{127}}$ etc.
Students must remember that when transferring the numbers or variable to the other side, the signs of the same will be changed to the opposite sign.