1. Equation

I have two questions:

The first is that I can't seem to get the answer to the attached equation. The book says x = 1, or x = 1/2. Though this is probably because of my second question:

The second question is, is there a method to work out how to factorise? Sometimes it seems impossible by looking at the numbers. Is there more than one method to do this?

Thanks.

I have two questions:

The first is that I can't seem to get the answer to the attached equation. The book says x = 1, or x = 1/2. Though this is probably because of my second question:

The second question is, is there a method to work out how to factorise? Sometimes it seems impossible by looking at the numbers. Is there more than one method to do this?

Thanks.
Let me just go through that working again:

$\displaystyle \frac{1}{x+1}-\frac{1}{2x+1}-\frac{1}{6}=0$

$\displaystyle 6(2x+1)-6(x+1)-(x+1)(2x+1)=0$

$\displaystyle 12x+6-(6x+6)-(2x^2+x+2x+1)=0$

$\displaystyle 12x+6-6x-6-2x^2-x-2x-1=0$

$\displaystyle -2x^2+3x-1=0$

$\displaystyle 2x^2-3x+1=0$

$\displaystyle 2x^2-x-2x+1=0$

$\displaystyle x(2x-1)-(2x-1)=0$

$\displaystyle (x-1)(2x-1)=0$

$\displaystyle x=1$ or $\displaystyle x=\frac{1}{2}$

Some extra info:
If you look at the discriminant, $\displaystyle \Delta=\sqrt{b^2-4ac}$, of a quadratic expression $\displaystyle ax^2+bx+c$, you can work out:
1. How many solutions it has
2. Whether or not it can be factored

For 1.
If $\displaystyle \Delta>0$, there are 2 real solutions.
If $\displaystyle \Delta = 0$, there is 1 real solution.
If $\displaystyle \Delta < 0$, there are no real solutions.

For 2.
If $\displaystyle \Delta$ is a perfect square, that is, if $\displaystyle \Delta=n^2,n \in Z$, then the expression can be factored, else it can't be.

For your original incorrect expression $\displaystyle 2x^2-6x+1=0$, $\displaystyle \Delta=\sqrt{(-6)^2-4 \times 2 \times 1}=\sqrt{28}$, and since 28 isn't a perfect square it can't be factorised.

Whereas $\displaystyle 2x^2-3x+1$ has discriminant $\displaystyle \Delta = \sqrt{(-3)^2-4 \times 2 \times 1}=\sqrt{1}$, which is a perfect square so it can.

There are other methods of finding roots, the first is completing the square, and the second is the quadratic formula:

The quadratic formula is very useful and you probably should remember it. For any expression $\displaystyle ax^2+bx+c=0$, the solutions are

$\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-b \pm \Delta}{2a}$

3. Completing the square

In attached I don't understand the the last line. Could you explain it please?

How can we replace the first 2 terms in that way? I can't quite see it.

Thank you.

4. $\displaystyle x^2+6x$

$\displaystyle =x^2+6x+\left(\frac{6}{2}\right)^2-\left(\frac{6}{2}\right)^2$

$\displaystyle =(x+3)^2-9$

To form a square with the first two terms of a quadratic expression, add and subtract the square of half the coefficient of the x term (line 2). Then factorise (line 3).

In general:

$\displaystyle ax^2+bx$

$\displaystyle =ax^2+bx+\left(\frac{b}{2} \right)^2-\left(\frac{b}{2} \right)^2$

$\displaystyle =\left(a+\frac{b}{2} \right)^2-\left(\frac{b}{2} \right)^2$