So, I don't get why that equals zero. Wouldn't that centre equation be -3? X can't equal all those numerals...
We have a test next week and the teacher didn't take time to make sure I understood.
That is because when you multiply two things and get zero, one or both of the numbers must be zero. Then you set up two equations and solve each one, like it says in your picture.
Souvik is completely correct. Think about it this way: A*0 = 0. A*B*C*D*E*F*0 = 0. Agreed? Now, say you have:
(x-a)(x-b) = 0
For this to equal zero, by the same logic, either (x-a) = 0 or (x-b) = 0 (or both), right? So x-a=0 -> x=a is one possibility, or x-b=0 ->x=b is the other possibility. Now we'll try plugging them in. x=a on the top line, x=b on the bottom line:
(a-a)(a-b) = 0*(a-b) = 0
(b-a)(b-b) = (b-a)*0 = 0
Both seem correct, agree? And the more parentheses you have, the more possible X values.
But, when you substitute 0 for X in (x-1)(x+3) you get (0-1)(0+3), and when you multiply the -1 and 3 by 'FOILing,' you get -3...
comicIDIOT wrote:
But, when you substitute 0 for X in (x-1)(x+3) you get (0-1)(0+3), and when you multiply the -1 and 3 by 'FOILing,' you get -3...
But zero isn't a solution to (X-1)(X+3) = 0, is it? X+3 = 0 or X-1 = 0 are solutions, ie, X={-3,1} are the solutions.
So, I'm basically using the wrong method to solve the equation?
comicIDIOT wrote:
So, I'm basically using the wrong method to solve the equation?
I'm not sure what method you're using. Say I give you this and tell you to solve for X; what do you do?
(X+5)(X-1)(X+2) = 0
I can't say how I'd do that.
But from what I've read above X can equal -5, +1 or -2; {-5,1,-2}.
In the prior problem, I was FOILing: First Outer Inner Last. So, (x-1)(x+3) would be x^2+2x-3 which would be 0^2+2(0)-3.
Correct on the first part, wrong on the second. x^2+2X-3 would not be 0^2+2(0)-3, because X=0 was not a solution. The whole EQUATION = 0 was a solution, in other words x^2+2x-3 = 0.
x^2+2x-3 = 0 where x=1: (1)^2+2(1)-3 = 1+2-3 = 0 (checks out)
x^2+2x-3 = 0 where X=-3: (-3)^2 + 2(-3)-3 = 9-6-3 = 0 (checks out)
So, we don't "FOIL" for these equations then? I was under the impression we were suppose to since the previous section (5.6) was about FOILing & Grouping, mainly. Now, we have this in 5.7.
comicIDIOT wrote:
So, we don't "FOIL" for these equations then? I was under the impression we were suppose to since the previous section (5.6) was about FOILing & Grouping, mainly. Now, we have this in 5.7.
The solutions are equally accurate whether you foil or don't, but it's easier to solve them if you convert to parenthetical form. Wouldn't you agree it's easier to see that -3 is a root from (x-1)(x+3)=0 than from x^2+2x-3=0?