The sum of a number and its positive square root is \(\frac{6}{25}\).

1.	The sum of a number and its positive square root is \(\frac{6}{25}\). The number is :(a) 5 (b) \(\frac15\) (c) 25 (d) \(\frac{1}{25}\)
Answer» (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 = \(\frac1{25}.\)

The sum of a number and its positive square root is \(\frac{6}{25}\). The number is :(a) 5 (b) \(\frac15\) (c) 25 (d) \(\frac{1}{25}\)

Answer»

(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 = \(\frac1{25}.\)