Question

If a²+b² ∝ ab, prove that (a+b) ∝ (a-b)


Please do it in a easy method. ​

Answer 1

Given, a² + b² ∝ ab

Or, a² + b² = k. ab [Where, k is a non zero variation constant]

Or, $\frac{ {a}^{2} + {b}^{2} }{ab} = \: k$

Or, $\frac{ {a}^{2} + \: {b}^{2} }{2 \: ab} = \frac{k}{2}$

By using componendo and dividendo, we can say :

$\frac{ {a}^{2} + \: {b}^{2} + 2ab}{ {a}^{2} + {b}^{2} - 2ab} \: = \frac{k + 2}{k - 2}$

Or, $\frac{(a + b)^{2} }{ ({a - b)}^{2} } = m \:$

In the above procedure, m is a constant equals to (k+2)/(k-2).

So, (a+b) = √m (a-b)

Or, (a+b) ∝ (a-b) [Since, √m is a non-zero variarion constant]

Hence, (a+b) ∝ (a-b) [PROVED]

Answer 2

Answer:

Step-by-step explanation: