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 :)
Printable View
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).Quote:
Originally Posted by 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?
Zero times any (finite) number is zero. This means that at least one of the factors on the LHS must be zero. Hence eitheror
.
Thank-you for that. Could you just do this one as I think I am on the right lines;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. Hence either
3a(a-1)=0
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.Quote:
Originally Posted by richard1234
So how would I do 3a(a-1)=0 if I had to solve it. The same with x^2-6x=0
As has been said in post #4,Quote:
Originally Posted by RonnieSaha
Quote:
Originally Posted by richard1234
I understand what the process is but I don't understand how to apply it to my equations
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.
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).
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.Quote:
Originally Posted by RonnieSaha
xa^2 - ya = 0
a(xa - y) = 0 : this is the "factoring" step
now, a = 0 and/or xa - y = 0 ; a = y/x
I understand that the 3 parts you suggested are a, b and c. However I don't know how to solve the quadratic anymore?
If you are talking about 3a(a - 1), I did not mentioned b and c. They are not part of the problem. The three factors in question are 3, a and a - 1. To continue, you have to say what follows from the fact that the product of these three numbers equals zero.Quote:
Originally Posted by RonnieSaha