1. ## Why?

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?

2. Hello,
Originally Posted by helloying
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?
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

3. 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.

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,

$x^{2} - 5x = 0$

$x - 5 = 0$ (dividing by $x$)

$x = 5$

The solution $x = 0$ has been lost.

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

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

Bring everything to one side, and put $RHS = 0.$

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

$3x^2 + 12x - x^2 - x = 0$

$2x^2 + 11x = 0$

Now FACTORISE:

$x (2x + 11) = 0$

Either $x = 0$ or $2x + 11 = 0 \Rightarrow x = \frac {-11}{2}$.

I hope that helps.

me07.