Help on homework!(Quadratics)

Jul 2012
10
0
A place
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 :)
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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 2012
10
0
A place
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?
 
Jun 2012
616
115
AZ
\(\displaystyle x(x-7) = 0\)

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 2012
10
0
A place
\(\displaystyle x(x-7) = 0\)

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 \(\displaystyle x-7 = 0 \Rightarrow x = 7\).
Thank-you for that. Could you just do this one as I think I am on the right lines;

3a(a-1)=0
 

emakarov

MHF Hall of Honor
Oct 2009
5,577
2,017
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 2012
10
0
A place
So how would I do 3a(a-1)=0 if I had to solve it. The same with x^2-6x=0
 
Jul 2012
10
0
A place
I understand what the process is but I don't understand how to apply it to my equations
 

emakarov

MHF Hall of Honor
Oct 2009
5,577
2,017
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.