Question

if x+y=5 and xy=6. find the value of x²+y² and x-y​

Answer 1

Answer:

given

x+y = 5

xy = 6

Now...

(x+y)² = x²+y²+2xy ===> 5² = x²+y²+12 ==> x²+y²= 13

there is an equation

(x+y)²= (x-y)²+4xy

on substitution

25 = (x-y)² +24 ===> (x-y)²= 1 ===> x-y = 1

Answer 2

Step-by-step explanation:

x + y = 5 - (i)

xy = 6

x = 6/y - (ii)

putting x = 6/y in eq (i)

6/y + y = 5

6 + y^2 / y = 5

y^2 + 6 = 5y

y^2 - 5y + 6= 0

y^2 - 3y - 2y + 6 = 0

y ( y - 3 ) - 2 ( y - 3 ) = 0

( y - 2) ( y - 3 ) = 0

therefore, y = 2 or 3

putting y = 2 in eq (ii)

x = 6/2

x = 3

putting y = 3 in eq (ii)

x = 6/3

x = 2

therefore x = 3 or 2

# case 1.

when x = 3 and y = 2

x^2 + y^2 = 3 ^2 + 2 ^2

= 9 + 4

= 13

x - y = 3-2 = 1

# case 2.

when x = 2 and y = 3

x^2 + y^2 = 2^2 + 3^2 = 13

x - y = 2 - 3 = -1