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

- Sep 26th 2008, 11:47 PM #1

- Sep 26th 2008, 11:51 PM #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

- Sep 27th 2008, 01:28 AM #3

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- Sep 2008
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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.

- Sep 27th 2008, 04:44 AM #4

- Joined
- Aug 2008
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- Dubai, UAE
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## 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,

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

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

$\displaystyle x = 5 $

The solution $\displaystyle x = 0 $ 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**

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

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

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

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

$\displaystyle 2x^2 + 11x = 0 $

Now FACTORISE:

$\displaystyle x (2x + 11) = 0 $

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

I hope that helps.

me07.