Question

If x + y = 3, x^2 + y^2 = 5 then xy is

Answer 1

x+y=3

x=3-y

putting the value of x

in this

x²+y²=5

(3-y) ²+y²=5

9-6y+y²+y²=5

2y²-6y+9-5=0

2y²-6y+4=0

2y²-4y-2y+4=0

2y(y-2)-2(y-2)=0

(2y-2) (y-2) =0

2y-2=0 or y-2=0

y=1 or y=2

so x=2 or 1

so x×y=2×1 or 1×2=2

I HOPE THIS WILL HELP YOU MARK ME BRAINLIEST

Answer 2

the given

$x + y = 3 \\ {x}^{2} + {y}^{2} = 5$

let x+y =3. ----------- equation 1

x^2 +y^2 = 5 ------------- equation 2

taken equation 1

$x + y = 3 \\ x = 3 - y$

x=3-y put in equation 2

${x}^{2} + {y}^{2} = 5 \\ {(3 - y)}^{2} + {y}^{2} = 5 \\ {3}^{2} + {y}^{2} - 2 \times 3 \times y = 5 \\ 9 + {y}^{2} - 6y = 5 \\ {y }^{2} - 6y + 9 - 5 = 0 \\ {y}^{2} - 6y + 4 = 0$

shri dhara chaerye method

$\frac{ - b \frac{ + }{ - } \sqrt{ {b}^{2} - 4ac} }{2a}$

$\frac{ 6 \frac{ + }{ - } \sqrt{36 - 16} }{2} \\ \frac{6 \frac{ + }{ - } \sqrt{20} }{2} \\ \frac{6 \frac{ + }{ - } 2 \sqrt{5} }{2} \\ 3 \frac{ + }{ - } \sqrt{5} \\ y = 3 + \sqrt{5} \\ x = 3 - \sqrt{5} \\ then \\ xy = (3 + \sqrt{5} ) \times (3 - \sqrt{5)} \\ xy = {3}^{2} - { (\sqrt{5}) }^{2} \\ xy = 9 - 5 \\ xy = 4 \: answer$