Question

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

Answer 1

Let the number be a

According to question

a + a^2 = 6 / 25

=> 25a^2 +25a = 6

=> 25a^2 + 25a - 6 = 0

=> 25a^2 +30a - 5a - 6 = 0

=> 5a(5a +6) - 1(5a+6) = 0

=> (5a-1)(5a+6) = 0


a = 1/5 and - 6/5

Neglecting negative value,

a = 1/5

Required number = 1/5

Answer 2

let the number be x.

By the given condition.

$x + \sqrt{x} = \frac{6}{25}$

$\sqrt{x} = \frac{6}{25} - x$

(by squaring both sides)

${( \sqrt{x} })^{2} = ( { \frac{6}{25} - x })^{2}$

we know that
$( {a - b)}^{2} = {a}^{2} - 2ab + {b}^{2}$
similarly,

$x = ( {\frac{6}{25} )}^{2} - 2( \frac{6}{25})(x) + {x}^{2}$

$x = \frac{36}{625} - \frac{12}{25} x + {x}^{2}$

(multiplying both sides by 625)

$625x = 36 - 300x + 625 {x}^{2}$

$625 {x}^{2} - 925x + 36 = 0$

by solving quadratic equation by factorization method

$625 {x}^{2} - 900x - 25x + 36 = 0$

$25x(25x - 36) - 1(25x - 36) \\ = 0$

$(25x - 36)(25x - 1) = 0$


we get,
25x-36=0 or 25x-1=0

$x = \frac{36}{25} \: or \: x = \frac{1}{25}$

hope this helps