Question

The sum of a number and its positive square root is $\frac{6}{25}$. Find the number.

Answer 1

Let the number be $x^2$.

Therefore, according to the question,

$\Rightarrow x^2+\sqrt{x^2}=\frac{6}{25}$

$\Rightarrow x^2+x=\frac{6}{25}$

Adding, $\frac{1}{4}$ on both the sides, we get,

$\Rightarrow x^2+x+\frac{1}{4}=\frac{6}{25}+\frac{1}{4}$

$\Rightarrow x^2+2\times \frac{1}{2}\times x+(\frac{1}{2}{)}^2=\frac{6\times 4+25}{100}$

$\Rightarrow (x+\frac{1}{2}{)}^2=\frac{49}{100}$

$\Rightarrow (x+\frac{1}{2})=\pm \frac{7}{10}$

$\Rightarrow x=\frac{7}{10}-\frac{1}{2}$ or $x=-\frac{7}{10}-\frac{1}{2}$

$\Rightarrow x=\frac{2}{10}$ or$\frac{-12}{10}$

Since, $x$ is the positive square root, therefore, $x=\frac{2}{10}$ and hence,

the required number is $(\frac{2}{10}{)}^2=\frac{1}{25}$