Question

Find the Value of x²+y² if i) x+y=5 and xy=6
ii) x-y=6 and xy=16​

Answer 1

Given the equation

x+y=5 and xy=6

Find out the value of (x³-y³) is

To proof

As given the equation

y = 6/5

put this in the x+y=5 equation

we get

x + \frac{6}{x} =5 x^{2} -5x +6 =0\\ x^{2} - 2x-3x+6 =0\\x(x-2)-3(x-2)=0x+x6=5x2−5x+6=0x2−2x−3x+6=0x(x−2)−3(x−2)=0

thus

(x -2) (x -3) =0

x=2,x=3

When x=2

put in x+y=5

y = 3

(x³-y³) = 8 - 27

= -19

when

x=3

put in x+y=5

(x³ -y³) = 27-8

= 19

Hence proved

Answer 2

Step-by-step explanation:

i)×^2+y^2+2×x×y-2×x×y (according to the identity)

(x+y)^2-2xy

according to the question,

x+y=5 and xy=6

therefore,

(x+y)^2=5^2=25

2xy=2×6=12

so the anwer is,25-12=13

ii)x^2+y^2-2xy+2xy(according to the identity)

therefore,

(x-y)^2=6^2=36

2xy=2×16=32

so the anwer is,36-32=4

Hope this is helpful