Question

sum of a number and its positive square root is 6/25.find the number

Answer 1

$let \: the \: number \: be \: x \\ x + \sqrt{x} = \frac{6}{25} \\ ( {x}^{2} ) \: \sqrt{x} - \frac{6}{25} = 0 \\ \sqrt{x} = - 1 \sqrt{( { - 1)}^{2} } ( - 4) \times 1( \frac{ - 6}{25}) \\ \sqrt{x} = - 1 + ( \frac{7}{5} ) \div 2 \\ = \frac{1}{5} - \frac{6}{5} \\ \\ \sqrt{x} = \frac{1}{5} \: ( {x}^{2} > 0) \\ x = \frac{(1}{5} {)}^{2} = \: \frac{1}{25}$

Answer 2

Here's your answer......