Question

X+y=5,xy=6,x>y
2(x2+y2)=?
X3_y3_3(x2+y2)=?
X5+y5=?

Answer 1

Answer:

Step-by-step explanation:

There are different methods to solve this question, one is given below

As we have given two equations:

x + y = 5 ———(1)

xy = 6 ————(2)

Now find “x” or “y” from equation (1)

x = 5 - y

now put this value in equation (2)

(5 - y )(y) = 6

5y - y^2 = 6

OR

-y^2 + 5y - 6 = 0

Multiplying both sides with “-1”

y^2 - 5y + 6 = 0 ——-(A)

Now using mid term break method. In this method we split/break the mid term into such two numbers whose sum is that mid term number while product is the product of first and last term.

Now guessing such two numbers whose product is equal to 6y^2 and sum is equal to -5y, so the two numbers are -3y and -2y ( -3y*-2y=6y^2 and -3y-2y=-5y)

so equation (A) becomes;

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

(y^2 - 3y) + (-2y + 6) = 0 ——- (making sets of two numbers having some terms in common)

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

taking y-3 common

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

using factor theorem we get

y - 3 = 0 & y - 2 = 0

so y = 3/2

Now putting value of y in equation (1)

x + 3 = 5

x = 5–3

x = 2

and

x + 2 = 5

x + 2 = 5

x = 5-2

x = 3