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 :)

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- Jul 24th 2012, 09:57 AMRonnieSahaHelp on homework!(Quadratics)
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 AMHallsofIvyRe: Help on homework!(Quadratics)
- Jul 24th 2012, 10:20 AMRonnieSahaRe: Help on homework!(Quadratics)
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 AMrichard1234Re: Help on homework!(Quadratics)
$\displaystyle x(x-7) = 0$

Zero times any (finite) number is zero. This means that at least one of the factors on the LHS must be zero. Hence either $\displaystyle x = 0$ or $\displaystyle x-7 = 0 \Rightarrow x = 7$. - Jul 24th 2012, 10:59 AMRonnieSahaRe: Help on homework!(Quadratics)
- Jul 24th 2012, 11:05 AMemakarovRe: Help on homework!(Quadratics)
- Jul 24th 2012, 11:27 AMRonnieSahaRe: Help on homework!(Quadratics)
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 AMemakarovRe: Help on homework!(Quadratics)
- Jul 24th 2012, 11:43 AMRonnieSahaRe: Help on homework!(Quadratics)
I understand what the process is but I don't understand how to apply it to my equations

- Jul 24th 2012, 11:53 AMemakarovRe: Help on homework!(Quadratics)
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 PMRonnieSahaRe: Help on homework!(Quadratics)
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 PMemakarovRe: Help on homework!(Quadratics)
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 PMWilmerRe: Help on homework!(Quadratics)
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 PMRonnieSahaRe: Help on homework!(Quadratics)
I understand that the 3 parts you suggested are a, b and c. However I don't know how to solve the quadratic anymore?

- Jul 24th 2012, 12:40 PMemakarovRe: Help on homework!(Quadratics)