Question

if x+y=12 and xy=7 then find x and y

Answer 1

in this pic your answer is given



x=11.385 (approx)
y=0.615 (approx)

Answer 2

Answer:

x = 6 ± √29

y = 6 ∓ √29

Step-by-step explanation:

Given,

x + y = 12  ( Let this be eqn 1 )

xy = 7 .......... eqn 2

From eqn 1 we get,

y = 12 - x ......... eqn 3

Substituting value of y in eqn 2 ,

x ( 12 - x ) = 7

12x - $x^{2}$ = 7

12x - $x^{2}$ - 7 = 0

Multiplying this equation with -1 ,

$x^{2}$ - 12x + 7 = 0

Solving the quadratic equation using quadratic formula :

( Quadratic formula : If the given equation is in form of

a$x^{2}$ + bx + c = 0 , the values of x that are solution of the equation is given by :

x =  $\frac{- b ± \sqrt{b ^ {2} - 4ac }}{2}$

on comparing above equation with a$x^{2}$ + bx + c = 0,

we get a= 1, b= -12 , c = 7

Substituting these values in quadratic formula :

x =  $\frac{- (-12) ± \sqrt{12 ^ {2} - 4 (1 ) ( 7 ) }}{2}$

x =  $\frac{ 12 ± \sqrt{ 144 - 28 }}{2}$

x = 6 ± √29

Hence, the value of x = 6 ± √29

Substituting value of x in eqn 3 :

y = x - 12

y = 12 - (6 ± √29)

y = 6 ∓ √29