Question

difference of two numbers is 2 and the sum of their square is 34 find the numbers​

Answer 1

let the 2 numbers be x and y

then,

x-y=2

x²+y²=34 (first equation)

x-y=2

x=2+y (second equation)

from first and second equation

(2+y)²+y²=34

4+y²+4y+y²=34

2y²+4y-30=0

2y²+10y-6y-30=0

2y(y+5) -6(y+5) =0

(2y-6)(y+5)=0

y=3, -5

therefore,

x=5, -3

Answer 2

Step-by-step explanation:

Let the two numbers be x and y.

its given that x-y=2(1)

and x^2+y^2=34(2)

From 1 we get,

x=y+2(3)

Substitute (3) in (2)

we get,

(y+2)^2+y^2=34

y^2+4y+4+y^2=34

2y^2+4y-30=0

Divide everything by 2,we get,

y^2+2y-15=0

By splitting the middle term,we get,

y^2-3y+5y-15=0

y(y-3)+5(y-3)=0

therefore,(y-3)(y+5)=0

y=3

Substitute the value in Equ3,we get

x=3+2

x=5

Therefore the numbers are 3 and 5.

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