How do you solve \[{{\left( 20-r \right)}^{\dfrac{1}{2}}}=r\]?

Hint: Square both the sides of the given equation and form a quadratic equation in r. Use the middle term split method to solve for the values of r. Reject the invalid value of r, i.e., the value of r that makes the function undefined. Consider the square root present in the original function as the positive square root.

Complete step by step answer:
Here, we have been provided with the function \[{{\left( 20-r \right)}^{\dfrac{1}{2}}}=r\] and we are asked to solve this equation. That means we have to find the value of x.
Now, we can see that in the L.H.S. of the given equation we have the expression of square root which must be removed to solve the equation. So, on squaring both the sides of the given equation, we have,
\[\Rightarrow \left( 20-r \right)={{r}^{2}}\]
Taking all the terms to the L.H.S., we have,
\[\Rightarrow 20-r-{{r}^{2}}=0\]
Multiplying both the sides with (-1), we have,
\[\Rightarrow {{r}^{2}}+r-20=0\]
Now, let us apply the middle term split method to solve the above quadratic equation. So, splitting the middle term into two terms such that their sum is r and product is equal to the product of the constant term (-20) and \[{{r}^{2}}\], i.e., \[-20{{r}^{2}}\], we get,
\[\begin{align}
  & \Rightarrow {{r}^{2}}+5r+\left( -4r \right)-20=0 \\
 & \Rightarrow {{r}^{2}}+5r-4r-20=0 \\
 & \Rightarrow r\left( r+5 \right)-4\left( r+5 \right)=0 \\
\end{align}\]
Taking \[\left( r+5 \right)\] common, we get,
\[\Rightarrow \left( r+5 \right)\left( r-4 \right)=0\]
Substituting each term equal to 0, we get,
\[\Rightarrow \left( r+5 \right)=0\] or \[\left( r-4 \right)=0\]
\[\Rightarrow r=-5\] or \[r=4\]
Here, we have obtained two values of r and one of them is negative. Now, as we can see that in the L.H.S. of the original equation it was the expression \[{{\left( 20-r \right)}^{\dfrac{1}{2}}}\], so when it will be solved for r = -5 it will give the answer \[\sqrt{25}=5\] and we know that \[-5\ne 5\]. So, it will lead us to the result \[{{\left( 20-r \right)}^{\dfrac{1}{2}}}\ne r\]. Therefore, we need to reject the value r = -5.

Hence, our answer is r = 4.

Note: One may note that here we have to consider only the positive square root, that means the number that will be obtained by solving the under-root term will always be positive. Hence, L.H.S. will always be positive and therefore R.H.S. cannot be negative. If the expression would have been like: - \[\pm \sqrt{20-r}=r\] then we would have considered r = -5 as our answer in addition to r = 4. Also, note that here you can apply the discriminant method to get the values of r but do not forget to reject the invalid value.