Solve for x from the given equation $ \dfrac{1}{2}\times 10\times \sqrt{{{x}^{2}}}-25=60 $ .

Hint: We will first simplify the LHS, $ \dfrac{1}{2}\times 10\times \sqrt{{{x}^{2}}}-25=60 $ as $ 5\sqrt{{{x}^{2}}}-25=60 $ and then in the second step, we will transpose -25 from LHS to RHS and solve it further. We will use the concept of exponent in last step, that we can write the square root of a term as, the term to the power $ \dfrac{1}{2} $ and here in question we have $ \sqrt{{{x}^{2}}} $ so, it can be written as $ {{\left( {{x}^{2}} \right)}^{\dfrac{1}{2}}} $ from this, 2 and $ \dfrac{1}{2} $ in power of x will get cancelled out and hence we get x with the power 1.

Complete step-by-step answer:
It is given in the question to solve for x. The given equation is –
 $ \dfrac{1}{2}\times 10\times \sqrt{{{x}^{2}}}-25=60 $
We will simplify LHS first to get
 $ 5\sqrt{{{x}^{2}}}-25=60 $
We now transpose all the constant part one side and variable part( here the term containing ‘x’ ) one side to solve the equation further, on transposing -25 from LHS to RHS, we get-
 $ 5\sqrt{{{x}^{2}}}=60+25 $
Solving right side, we get,
 $ 5\sqrt{{{x}^{2}}}=85 $
On dividing both the sides with 5, we get –
\[\dfrac{5\sqrt{{{x}^{2}}}}{5}=\dfrac{85}{5}\]
Solving further, we get
 $ \sqrt{{{x}^{2}}}=17 $
Now, we know that square root can be considered as a power of $ \dfrac{1}{2} $ , therefore, replacing the value of under root as $ \dfrac{1}{2} $ , we get, \[{{\left( {{x}^{2}} \right)}^{\dfrac{1}{2}}}=17\].
Therefore, multiplying the power, that is, multiplying $ \dfrac{1}{2} $ and 2, we get $ \dfrac{1}{2}\times 2=1 $ .
Hence, power of x becomes 1 and we get $ x=17 $ .
Therefore, the value of x for the given equation will be $ x=17 $ .

Note: It is noted that most of the students make mistakes in taking sign while transposing the term from one side to another. Also, some students may miss the - sign and then they might start cancelling factors from terms like 10 and 60 on both sides. So, they must be careful of such mistakes while solving this question. Also, the students can directly write $ \sqrt{{{x}^{2}}} $ as x, as we know that the square root of any square is the term itself. For example, $ \sqrt{{{4}^{2}}}=4\text{ and }\sqrt{{{\left( x+y \right)}^{2}}}=\left( x+y \right) $ .