3x(x+4)=x(x+1)

so i divide both side by x

3(x+4)=(x+1)

then 3x+12=x+1

2x=-11

x=(-11/2)

but this is incorrect. why?

Results 1 to 4 of 4

- September 27th 2008, 12:47 AM #1

- September 27th 2008, 12:51 AM #2
Hello,

When solving this type of equation, you mustn't divide. You must factorise, because you miss information this way.

If you divide both sides by 1, you "miss" the solution x=0.

3x(x+4)=x(x+1)

3x(x+4)-x(x+1)=0

x(3x+12-x-1)=0

x(2x+11)=0

the two solutions are :

- x such that x=0

- x such that 2x+11=0

- September 27th 2008, 02:28 AM #3

- Joined
- Sep 2008
- Posts
- 12

Remember that you can divide by x only if x different from zero!

So in case you choose to divide by x, you must precise that x is different from zero.

And then to be thorough you need to examine what happens when x takes the value 0.

This is why it is safer not to divide but factorise.

- September 27th 2008, 05:44 AM #4

- Joined
- Aug 2008
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- Dubai, UAE
- Posts
- 58

## Quadratic equations

Quadratic equations sometimes have a common factor containing the unknown quantity. In such cases, dividing by the common factor leads to the loss of a solution.

For instance,

(dividing by )

The solution has been lost.

Although dividing an equation by a**NUMERICAL**common factor is correct and sensible, dividing by a common factorresults in the loss of a solution.**containing the unknown quantity**

Bring everything to one side, and put

Now FACTORISE:

Either or .

I hope that helps.

me07.