Question

if x+y=12 and xy=7find the value of xand y

Answer 1

x + y = 12          ------ : [ 1 ]

xy = 7               ------ : [ 2 ]

In equation 1,

x + y = 12

x = 12 - y          ------- : [ 3 ]

In equation 2,

xy = 7

( 12 - y ) y = 7          [ ∴ x = 12 - y ]

12y - y^2 = 7

y^2 - 12y + 7 = 0

√Discriminant = √{ b^2 - 4ac }

     

                      = √{ ( - 12 )^2 - 4( 1 × 7 ) }

                      = √{ 144 - 28 }

                      = √116

                      = 2√29       ------- : [ 4 ]

By Quadratic equation,

x = $\dfrac{ -b \pm \sqrt{discriminant}}{2a}$

  = $\dfrac{-( - 12 ) \pm 2\sqrt{29}}{2( 1 ) }$       [ From 4 ]

  = $\dfrac{12 + 2\sqrt{29}}{2}$

  = $\dfrac{2( 6 + \sqrt{29})}{2}$

x  = 6 + √29

Then, putting the value of x in [ 1 ]

x + y = 12

6 + √29 + y = 12

y = 12 - 6 - √29

y = 6 - √29

Answer 2

Given, x + y = 12

x = 12 - y


Given, xy = 7

( 12 - y )y = 7

12y - y² = 7

0 = 7 + y² - 12y

0 = y² - 12y + 7



By Quadratic formula,


x =$\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2}$


x =$\dfrac{12\pm \sqrt{( - 12 )^{2}-4( 1 \times 7 )}}{2 \times 1}$


x =$\dfrac{12 \pm \sqrt{116}}{2}$


x =$\dfrac{ 2( 6 \pm \sqrt{29}}{2}$


x = 6 + √29 or 6 - √29




Therefore, talking positive value

x + y = 12

6 + √29 + y = 12

y = 6 - √29





Hence,

$x = 6 + \sqrt{29} \\ y = 6 - \sqrt{29}$