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• Jul 24th 2012, 09:57 AM
RonnieSaha
Hello...

I have this problem on my homework, I have to solve the quadratics by factorising;

x(x-7)=0

If you could explain how to do it (simple terms!) I would be grateful :)
• Jul 24th 2012, 10:04 AM
HallsofIvy
Quote:

Originally Posted by RonnieSaha
Hello...

I have this problem on my homework, I have to solve the quadratics by factorising;

x(x-7)=0

If you could explain how to do it (simple terms!) I would be grateful :)

That's already "factored". Now you need the "zero product rule": if ab= 0 then either a= 0 or b= 0 (or both).
• Jul 24th 2012, 10:20 AM
RonnieSaha
I really don't understand the process you have explained above... could you please go through the stages of solving x(x-7) with the zero product rule?
• Jul 24th 2012, 10:54 AM
richard1234
$x(x-7) = 0$

Zero times any (finite) number is zero. This means that at least one of the factors on the LHS must be zero. Hence either $x = 0$ or $x-7 = 0 \Rightarrow x = 7$.
• Jul 24th 2012, 10:59 AM
RonnieSaha
Quote:

Originally Posted by richard1234
$x(x-7) = 0$

Zero times any (finite) number is zero. This means that at least one of the factors on the LHS must be zero. Hence either $x = 0$ or $x-7 = 0 \Rightarrow x = 7$.

Thank-you for that. Could you just do this one as I think I am on the right lines;

3a(a-1)=0
• Jul 24th 2012, 11:05 AM
emakarov
Quote:

Originally Posted by richard1234
Zero times any (finite) number is zero. This means that at least one of the factors on the LHS must be zero.

The second claim is certainly true for real numbers, but it does not follow from the first one. In every ring we have 0 * x = 0, but in some rings x * y = 0 does not imply that x = 0 or y = 0.
• Jul 24th 2012, 11:27 AM
RonnieSaha
So how would I do 3a(a-1)=0 if I had to solve it. The same with x^2-6x=0
• Jul 24th 2012, 11:39 AM
emakarov
Quote:

Originally Posted by RonnieSaha
So how would I do 3a(a-1)=0 if I had to solve it.

As has been said in post #4,
Quote:

Originally Posted by richard1234
at least one of the factors on the LHS must be zero.

• Jul 24th 2012, 11:43 AM
RonnieSaha
I understand what the process is but I don't understand how to apply it to my equations
• Jul 24th 2012, 11:53 AM
emakarov
I would be willing to talk about this if in exchange you explain in a very detailed way what exactly you don't understand about this problem. I find it hard to believe that you don't know how to solve 3a(a - 1) = 0 after you've been shown how to solve x(x - 7) = 0 and have been told that a product is zero if and only if one of the factors is zero. It may be that you don't know what a factor is or you don't know how to solve linear equations. I'd like to know what your reasoning is before further explanations.
• Jul 24th 2012, 12:05 PM
RonnieSaha
Okay... I can see that the left hand side must equal to the right hand side which is 0. I don't know how you get from the quadratic to the point of finding out what x or a could be. I don't know what the factors are and so don't understand where to put the 0 in either a or b. In fact, i don't know where a or b are in x(x - 7) or 3a(a - 1).
• Jul 24th 2012, 12:15 PM
emakarov
Quote:

Originally Posted by RonnieSaha
I don't know what the factors are

Ah, now we are talking. The left-hand side, 3a(a - 1), is a product of three numbers: 3, a and a - 1. Here a is some number that we don't know yet and have to find. Each of those three numbers, i.e., 3, a and a - 1, are called factors. Now, multiplication has this property that if the product equals zero, then one or more of the factors also equal zero.
• Jul 24th 2012, 12:18 PM
Wilmer
xa^2 - ya = 0
a(xa - y) = 0 : this is the "factoring" step

now, a = 0 and/or xa - y = 0 ; a = y/x
• Jul 24th 2012, 12:23 PM
RonnieSaha