Question

x + y = 42, xy = 108 then what is x and y. Is there any (non-quadratic)formula for finding this instead of the process of elimination

Answer 1

Answer:

when x = 39.25, y = 2.75

when x = 2.75, x = 39.25

Step-by-step explanation:

use substitution method

x+y = 42 ------ (1)

xy = 108 ------ (2)

from (1) ,

x+y = 42

y = 42-x ------ (3)

substitute (3) into (2)

xy = 108

x( 42-x ) = 108

42x - x² = 108

-x² + 42x - 108 = 0

x² - 42x + 108 = 0

use the following formula to solve the value of x:

`i) \: x = \frac{ - b + \sqrt{ {b}^(2) - 4ac } }{2a} \: \\ ii) \: x = \frac{ - b - \sqrt{ {b}^(2) - 4ac } }{2a}`

a = 1

b = -42

c = 108

`i) \: x = \frac{ - b + \sqrt{ {b}^(2) - 4ac} }{2a}`

`x = \frac{ - ( - 42) + \sqrt{( - 42) {}^(2) - 4(1)(108)} }{2(1)}`

`x = (42 + √(1764 - 432) )/(2)`

`x = (42 + √(1332) )/(2)`

`x = (42 + 6 √(37) )/(2)`

`x = 21 + 3 √(37)`

`x = 39.25`

`ii) \: x \: \frac{ - b - \sqrt{ {b}^(2) - 4ac} }{2a}`

`x = \frac{ - ( - 42) - \sqrt{( - 42) {}^(2) - 4(1)(108} }{2(1)}`

`x = (42 - √(1764 - 432) )/(2)`

`x = (42 - √(1332) )/(2)`

`x = (42 - 6 √(37) )/(2)`

`x = 21 - 3 √(37)`

`x = 2.75`

the 2 values of x are x = 39.25 and x = 2.75

substitute x = 39.25 into (3)

y = 42-x

y = 42-39.25

y = 2.75

substitute x = 2.75 into (3)

y = 42-x

y = 42-2.75

y = 39.25

when x = 39.25, y = 2.75

when x = 2.75, x = 39.25