Somethings go wrong...

Algebra Level 2

This problem was not created by me.

Let a = b.

$1.\quad { a }^{ 2 }=ab\\ { 2.\quad a }^{ 2 }-{ b }^{ 2 }={ a }^{ 2 }-ab\\ 3.\quad (a-b)(a+b)=a(a-b)\\ 4.\quad a+b\quad =\quad a\\ 5.\quad b\quad =\quad a\quad =\quad 0$

So all numbers are equal to 0... What step went wrong? (Try not to take more than 2 minutes.)

4 to 5 3 to 4 1 to 2 2 to 3

3 solutions

Abdur Rehman Zahid
Nov 13, 2014

If $a=b$ then obviously $a-b=0$ .In the $3rd$ step he factorised $a^2-b^2\;as\;(a+b)(a-b)$ and in the $4th$ step he cancelled out $a-b=\boxed{0}\;from\;both\;sides$ and chose the result as 1.But $\frac{0}{0}=Undefined$ so steps $\boxed{3\;to\;4}$ are wrong

Ankush Rathi
Aug 2, 2014

The equations 3 and 4 if considered contradicts our assumption that a=b.As explained below

We have (a-b)(a+b) =a(a-b) so a =a+b ie. Equation 4

Now consider eqn 4 a+b=a But given assumption is a=b So, the eqn becomes a=2b But a is not equal to 2b

Hence eqns 3 and 4 are wrong.

Bro how is the third one wrong?

Sourabh Nolkha - 6 years, 10 months ago

because he is cancelling a-b from both sides which means he is taking 0/0 = 1 but it is undefined

akash deep - 6 years, 10 months ago

Rick B
Aug 2, 2014

To get the fourth equation, the third one was divided by (a - b). Since a = b, we can write (a - b) as (a - a) or (b - b), which is obviously zero, meaning that the division was by zero.

This is wrong question.

Sourabh Nolkha - 6 years, 10 months ago

Please explain.

Tan Wee Kean - 6 years, 10 months ago

Somethings go wrong...

3 solutions

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