Question

prove that:. x²-y²= (x+y)(x-y)​

Answer 1

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To prove :

x²-y² =(x+y) (x-y)

Taking RHS;

(x+y) (x-y)

x(x-y) +y(x-y)

x²-xy+yx-y²

x²-y²

Hence, Proved.

Answer 2

Answer:

Consider the given equation

(x + y)∝(x — y)

If 'k' is any constant then

x + y = k(x — y)

(x + y)/(x — y) = k

Squaring on both sides we get

(x+y)²/(x — y)² = k²

(x² + y² +2xy)/(x²+y² — 2xy)=k²

Applying componendo and dividendo

(x²+y²+2xy+x²+y²—2xy)/(x²+y²+2xy—x²—y²+2xy) = (k²+1)/(k²—1)

Cancelling and simplifying

(2(x²+y²))/(4xy)=(k²+1)/(k²—1)

Cancelling common terms

(x² + y²)/(2xy) = (k²+1)/(k² — 1)

(x²+y²)/(xy)=(2(k²+1))/(k² — 1)

Since 'k' is proportionality constant the whole term in the right hand side is also a constant, this means say

(2(k²+1))/(k²—1)=C

So, this gives,

(x²+y²)/(xy)=C

x² + y² = Cxy

Now if y∝x ,if and only if y=my for any real constant 'm' so the above equation can be written as

x² + y² ∝ xy

This proves the result.