Question

44. The sum of a number and its positive

square root is 6/25 Find the number.​

Answer 1

Answer:

x = 36/25  

x = 1/25

Step-by-step explanation:

let the number be x

then its given that, x + √x = 6/25

=> √x²+√x = 6/25

=>  25√x²+25√x = 6

=> 25√x²+25√x - 6 = 0

let y = √x

=> 25y²+25y - 6 = 0

=> 25y²+30y-5y - 6 = 0

=> 5y(5y+6) - (5y+6) = 0

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

=> either 5y+6 = 0   or 5y-1 = 0

=> either 5y = -6   or 5y = 1

=> either y = -6/5   or y = 1/5

ie., either √x = -6/5   or √x = 1/5

ie., either x = (-6/5)²   or x = (1/5)²

ie., either x = 36/25   or x = 1/25

Answer 2

Answer :-

Let the number be x.

According to the question :-

$\implies\sf x + \sqrt{x} = \dfrac{6}{25}$

$\implies\sf \sqrt{x} = \dfrac{6}{25} - x$

Squaring on both side :-

$\implies\sf (\sqrt{x})^2 = \left ( \dfrac{6}{25} - x \right )^2$

$\implies\sf x = \dfrac{36}{625} + x^2 - 2 \times \dfrac{6}{25} \times x$

$\implies\sf x = \dfrac{36}{625} + x^2 - \dfrac{12}{25}x$

$\implies\sf x + \dfrac{12}{25}x = \dfrac{36}{625} + x^2$

$\implies\sf \dfrac{37}{25} x = \dfrac{36}{625} + x^2$

$\implies\sf x^2 + \dfrac{36}{625} - \dfrac{37}{25} x = 0$

$\implies\sf\dfrac{625x^2 - 925x + 36}{625} = 0$

$\implies\sf 625x^2 - 925x + 36 = 0$

Using quadratic formula :-

$\implies\sf Roots =\dfrac{ - ( -925 ) \pm \sqrt{925\times 925 - 4 \times 625 \times 36}}{2 \times 625}$

$\implies\sf \dfrac{925 \pm \sqrt{765625}}{1250}$

$\implies\sf \dfrac{925 + 875}{1250} , \dfrac{925 - 875}{1250}$

$\implies\sf 1.44 , 0.04$

Now, 1.44 is not possible because the question has given us the sum of a number and it's positive square root.

Required number = 0.04