someone please prove this wrong?

Question

25-55 = 36-66
(5)^2-(2*5*11/2) = (6)^2-(2*6*11/2)

Adding (11/2)^2 on both sides :

(5)^2-(2*5*11/2)+(11/2)^2=(6)^2-(2*6*11/2)+(11/2)^2
=>(5-11/2)^2=(6-11/2)^2

taking square root on both sides :
5-11/2=6-11/2
=> 5=6

Guys please disprove it mathematically and using logic i mean wot is the flaw or wot is the loop hole which has been misused?

I understamd tat by taking square root v get 4 cases
Case 1:When both the sides are positive
Case 2:When both sides are negative
Case 3:When Left side is positive and the Right side is negative
Case 4:When left side is negative and the right side is positive.
Now prove me wrong when both the sides are taken to b positive or negative

Answer 1

the fallacy is in the last step
(5-11/2)^2=(6-11/2)^2
when you take square root you get 5-11/2= plus or minus 6-11/2
in this case only the negative value is acceptable as otherwise you get the absurd answer of 5=6
so 5-11/2=-(6-11/2)
which is perfectly mathematically correct

Answer 2

The problem is in your last step.

When taking the square root of both sides of an equation, one has to remember that one side can be the negative of the other, because every positive quantity has two square roots: one positive, one negative.

For instance, it is true that 3² = (-3)²; both sides equal 9. However, if I take the square root of both sides carelessly, I get 3 = -3, which is obviously false. What I should write, instead, is 3 = - (-3), which is true.

In your case, it's *not* the case that 5 - 11/2 = 6 - 11/2, but it *is* the case that 5 - 11/2 = - (6 - 11/2). The left side comes to 10/2 - 11/2 = -1/2, and the right side comes to - (12/2 - 11/2) = - (1/2) = -1/2.

Answer 3

When you're on the line
(5 - 11/2)² = (6 - 11/2)²,
your math is correct... both sides clearly equal 1/4.

Your next line, however, is wrong. If you're going to take a square root of both sides of an equation, do it properly.
√(5 - 11/2)² = √(6 - 11/2)²
√(-1/2)² = √(1/2)²
√(1/4) = √(1/4)
1/2 = 1/2

Remember back in the first chapter of your algebra text when you were supposed to have learned about the proper order of operations? (Grouping symbols first, then exponents and roots, and so on.) Each radical is a grouping symbol and you must evaluate the square first, before doing the square root.

If you're insistent that you want to take the square root first, by taking the square root on the left, and try it with complex numbers, that doesn't work, either.
√(5 - 11/2)² = √(6 - 11/2)²
√(-1/2)² = √(1/2)²
√(-1)² · √(1/2)² = √(1/2)²
√(1)² · √(1/2)² = √(1/2)²
(1) (1/2) = (1/2)
(1/2) = (1/2)

Answer 4

(5-11/2) < 0. The result of a square root operation is always > 0, so when you take the square root of (5-11/2)^2 you answer should be -(5-11/2) = 11/2 - 5

Answer 5

only the last step is wrong - when you apply square root to something, the answer can be -ve or +ve. Eg: whats square of
(-2)? its 4. Whats square of 2? thats 4 too. So now, whats the square root of 4? +2 or -2. Apply the same in your case and choose the answer that makes sense. In your case you'd have to take the -ve.

Here's similar one:

Given: a = b

=> a^2 = ab..................since a = b
=> b^2 - a^2 = b^2 - ab...........subtracting the same value from b^2
=> (b-a)(b+a) = b(b-a).......factorising
=> b + a = b...........cancelling
But b = a
=> 2b = b
=> 2 =1

the step named 'cancelling' is wrong. Why? because (b-a) is zero and you can't cancel zero on both sides of the equation and say they're equal(eg: 0 x 5 = 0 x 7 , you can't cancel 0 and say 5 =7)

Answer 6

The reason this problem wouldn't work is because you can't take the square root of either sides until you simplify your work.(ie. can't take square root of 6+3, only square root of 9) However,in some cases you could use the square root function for special cases, like sqrt([x+3]^2) to get |x+3| . The order of operations and what you can do is very important.
If you click the link below it shows how people can make 0=1 because they did things you shouldn't do in math, such as divide by an unknown variable(it could be zero).

Answer 7

5 - 11/2 = 6 - 11/2 or -(6 - 11/2)
In this case, 5 - 11/2 = -(6 - 11/2) and not 6 - 11/2

Answer 8

The problem is when you took the square root.

sqrt (x ^ 2) = abs x

it does not equal x

At least thats the standard convention.

Ultimatly there are two roots to a positive number b^2 (which it is since its a square) either |b| or -|b|

The convention is that the root is |b|. You decided in one case to make it |b| and in another case -|b|. Since you did it inconsistantly you arrived at the wrong conclusion.

Answer 9

=>(5-11/2)^2=(6-11/2)^2
=>(-0.5)^2 = (0.5)^2

This is perfectly true.

In the sense that what you meant by "take the square root of both sides" is "remove the square root symbols from both sides"

You will get -0.5 = 0.5 a contradiction.

But mathematics requires you to take the square root of a number:

=>(5-11/2)^2=(6-11/2)^2
=>(-0.5)^2 = (0.5)^2
=> 0.25 = 0.25

You could now take the square root if you like.

Answer 10

sorry man,
im bad at math
but i know this
0+2= 2
so 2 pts...thanks
jajja