Question

how to prove 2+2=5 tricy question unable to find the answer help me guys

Answer 1

The answer is given below :

Let us consider

x = 1

Now, multiplying both sides by x, we get

x . x = 1 . x

=> x² = x

Now, subtracting 1 from both sides, we get

x² - 1 = x - 1

=> (x + 1)(x - 1) = x - 1

=> x + 1 = 1

When x = 1, we get

1 + 1 = 1

=> 2 = 1

Now, adding 3 to both sides, we get

2 + 3 = 1 + 3

=> 5 = 4

=> 4 = 5

So, 4 = 5. [Proved]

Thank you for your question.

Answer 2

Solution :

_____________________________________________________________

Given :

To Prove :

⇒ 2 + 2 = 5   : P

_____________________________________________________________

Proof :

⇒ We know that 2 + 2 = 4,.

So,.

⇒ 2 + 2 = 4

⇒ $2 + 2 = 4 - \frac{9}{2} + \frac{9}{2}$   ( Adding & subtracting $\frac{9}{2}$  ),.

⇒ $2 + 2 = \sqrt{(4 - \frac{9}{2} )^2} + \frac{9}{2}$

By taking square root & square for $4 - \frac{9}{2}$ ,.

⇒ $2 + 2 = \sqrt{ (4)^2 - 2(4)( \frac{9}{2} ) + ( \frac{9}{2} )^2} + \frac{9}{2}$

We know that,

⇒ (a - b)² = a² - 2ab + b²

Substituting $a = 4$ & $b = \frac{9}{2}$,

We get,

⇒ $2 + 2 = \sqrt{16 - 8( \frac{9}{2}) + ( \frac{9}{2} )^2 } + \frac{9}{2}$

⇒ $2 + 2 = \sqrt{16 - \frac{72}{2} + ( \frac{9}{2} )^2 } + \frac{9}{2}$

⇒ $2 + 2 = \sqrt{16 - 36 + (\frac{9}{2})^2 } + \frac{9}{2}$

⇒ $2 + 2 = \sqrt{25 - 45 + (\frac{9}{2})^2 } + \frac{9}{2}$

⇒ $2 + 2 = \sqrt{(5)^2 - 2(5)( \frac{9}{2} ) + ( \frac{9}{2})^2} + \frac{9}{2}$

It is again in the form,

a² - 2ab + b²,
(Where a = 5 & $b = \frac{9}{2}$)

So, we can factorize to the form (a - b)²,.

We get,

⇒ $2 + 2 = \sqrt{(5 - \frac{9}{2})^2 } + \frac{9}{2}$

By canceling square & square root,

We get,

⇒ $2 + 2 = 5 - \frac{9}{2} + \frac{9}{2}$

By canceling $\frac{9}{2}$ &$-\frac{9}{2}$

We get,

⇒⇒⇒ $2 + 2 = 5$ : P

_____________________________________________________________

                                Hope it Helps !!