(d) \(\frac1{25}\)
Let the number be x.
Then x + \(\sqrt{x}= \frac{6}{25}\)
⇒ 25x + 25\(\sqrt{x} = 6\)
Let \(\sqrt{x}\) = y. Then, the equation becomes
25y2 + 25y - 6 = 0
⇒ 25y2 + 30y - 5y - 6 = 0
⇒ 5y(5y + 6) - 1(5y + 6) = 0
⇒ (5y + 6)(5y - 1) = 0 ⇒ y = \(-\frac65,\frac15\)
Rejecting the negative value , \(\sqrt{x}\) = \(\frac15\) ⇒ x = \(\frac1{25}.\)