Question

the sum of two natural numbers is 28 and their product is 192 find the numbers ​

Answer 1

Answer:-

let's consider

first no = a

second no= b

a+b = 28

a= 28-b ---------------------(1)

product of no. = a×b = 192 --------------(2)

$= > (28 - b) \times b = 192 \\ = > 28b - {b}^{2} = 192 \\ = > 28b - {b}^{2} - 192 = 0 \\ = > {b}^{2} - 28b + 192 = 0 \\ = > {b}^{2} - (16 + 12)b + 192 = 0 \\ = > {b}^{2} 16b - 12b + 192 = 0 \\ = > b(b - 16) - 12(b - 16) = 0 \\ = > (b - 16)(b - 12) \\ = > b - 16 = 0 \\ = > b = 16 \\=> 12$