[Guest]

I came across this on the web, and i don't see what is wrong with it. Perhpas you guys could figure it out? I apologize if this isn't on the right forum section.

Suppose "a" and "b" are any 2 numbers. Also, a + b = t

(a+b)(a-b) = t(a-b)

a^2 - b^2 = ta-tb

a^2 -ta = b^2 - tb

a^2 -ta +(t/2)^2 = b^2 -tb +(t/2)^2

(a-t/2)^2 = (b-t/2)^2

By taking the square root of both sides.

a-t/2 = b-t/2

a = b-t/2+t/2

a=b

Since a=b, all numbers are the same, and math is therefore pointless.

[Prime]

Square root has two solutions -- positive and negative. x² = y² does not mean x = y.

Case in point, your equation before taking square root:

((a-b)/2)² = ((b-a)/2)² perfectly okay at that point.

Whereas, (a-b)/2 = (b-a)/2 is no longer true, unless a was equal to b to start with.

[peace*out]

And HOW does that work??

[Izzy]

And HOW does that work??

Plug in any two numbers for a and b, and solve the equations. I'll use 2 and 5 as an example:

(a+b)(a-b) = t(a-b)

(2+5)(2-5) = 7(2-5)

(7)(-3) = 7(-3)

-21 = -21

a^2 - b^2 = ta-tb

2^2 - 5^2 = 7(2) - 7(5)

4 - 25 = 14 - 35

-21 = -21

I'm too lazy to do the rest, but you get the idea.

Just because they both equal -21 does NOT make maths pointless.

[peace*out]

ohhhhhhhhhhhhhh........

thanks izzy!!!

[Guest]

From an algebra stand point, the reason the equation works out to "a=b" is because t is removed.

Algebra Rule number 4867235: Every time you divide by or eliminate a variable, 1 or more of the solutions to your problem goes away.

Hope this helps

Edit: Clarification

Edited October 20, 2008 by pw0nzd

[Guest]

It is the same as saying

9 = 9

3² = -3²

By taking the square root of both sides.

3 = -3

Well, that is wrong

I came across this on the web, and i don't see what is wrong with it. Perhpas you guys could figure it out? I apologize if this isn't on the right forum section.

Suppose "a" and "b" are any 2 numbers. Also, a + b = t

(a+b)(a-b) = t(a-b)

a^2 - b^2 = ta-tb

a^2 -ta = b^2 - tb

a^2 -ta +(t/2)^2 = b^2 -tb +(t/2)^2

(a-t/2)^2 = (b-t/2)^2

By taking the square root of both sides.

a-t/2 = b-t/2

a = b-t/2+t/2

a=b

Since a=b, all numbers are the same, and math is therefore pointless.

[Guest]

Square root has two solutions -- positive and negative. x² = y² does not mean x = y.

Case in point, your equation before taking square root:

((a-b)/2)² = ((b-a)/2)² perfectly okay at that point.

Whereas, (a-b)/2 = (b-a)/2 is no longer true, unless a was equal to b to start with.

Prime is right

(a+b)(a-b) = t(a-b)

a^2 - b^2 = ta-tb

a^2 -ta = b^2 - tb

a^2 -ta +(t/2)^2 = b^2 -tb +(t/2)^2

(a-t/2)^2 = (b-t/2)^2

to meet the above condition, (a-t/2) = plus /minus (b-t/2)

So: a-t/2 = b-t/2 or a-t/2 = -(b-t/2)

a = b-t/2+t/2 or a-t/2=t/2-b

a=b or a+b = t

So, to meet the condition, either a=b or a and b are having different number but add up to become t !

[bonanova]

If a = b your first equation is meaningless and you can go basically anywhere you like with it:

(a+b) * 0 = t * 0 = 56t * 0 = t³ * 0 = ...

so you could conclude

(a+b) = t = 56t = t³ = ... you get the idea.

In general, beware of equations that have a-b in them.