Question

The sum of a number and its positive square root is $\dfrac{6}{{25}}$. The number is:
A. 5
B. $\dfrac{1}{5}$
C. 25
D. $\dfrac{1}{{25}}$

Answer 1

Hint: We can take the number as a variable. Then we can form an equation with the given details. After simplification and substitution, we get a quadratic equation. We can solve the quadratic equation to get the solutions and the non-acceptable solutions are rejected. Then on resubstituting for our substitution, we get the required number.

Complete step by step Answer:

Let us assume the be $x$. Then we can write its square root as \[\sqrt x \].
We are given that sum of the number and positive square root is $\dfrac{6}{{25}}$. We can make this into a mathematical equation.
$ \Rightarrow x + \sqrt x = \dfrac{6}{{25}}$
We can multiply the equation with 25.
$ \Rightarrow 25x + 25\sqrt x = 6$
We can give a substitution for \[\sqrt x \]. Let $y = \sqrt x $. Then ${y^2} = x$
$ \Rightarrow 25{y^2} + 25y - 6 = 0$
Now we have a quadratic equation, we solve it by the method of factorization.
On splitting the middle term we get,
$ \Rightarrow 25{y^2} + 30y - 5y - 6 = 0$
We can take the common terms,
$ \Rightarrow 5y\left( {5y + 6} \right) - 1\left( {5y + 6} \right) = 0$
Again, we can take the common terms,
$ \Rightarrow \left( {5y + 6} \right)\left( {5y - 1} \right) = 0$
On equating both the factors to 0, we get,
$y = - \dfrac{6}{5},\dfrac{1}{5}$
As y is a square root of a number, it cannot have negative value. So, we can reject $y = - \dfrac{6}{5}$
$ \Rightarrow y = \dfrac{1}{5}$
Now we can find x.
We have $y = \sqrt x $, we can take squares on both sides
$ \Rightarrow x = {y^2}$
On substituting the value of y, we get,
$ \Rightarrow x = {\left( {\dfrac{1}{5}} \right)^2}$
$ \Rightarrow x = \dfrac{1}{{25}}$
Therefore, the required number is $\dfrac{1}{{25}}$

Hence, the correct answer is option D.

Note: The concept of making mathematical equations from given statements is known as mathematical modeling. We used the method of substitution to get rid of the square root and to make the equation a quadratic equation. The quadratic equation can be solved by factorization or by using the quadratic formula. The solution will be the same for both methods. After getting the roots, the negative values are rejected. This is because we took y as the square root of a number. A square root cannot be negative for real numbers. So, y also cannot be negative. Hence, we reject the negative values of y. We must resubstitute use substitution to get the final required answer