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• Jul 24th 2012, 12:40 PM
Wilmer
Quote:

Originally Posted by RonnieSaha
I understand that the 3 parts you suggested are a, b and c. However I don't know how to solve the quadratic anymore?

NO. I was attempting to show a "general case" when the "C" = 0 (which is the case with your problem).

Let me do it another way:
Ax^2 + Bx + 0 = 0 : C=0; ok?
so: Ax^2 + Bx = 0
Now factor (no need for the quadratic):
x(Ax + B) = 0
so: x = 0 or Ax + B = 0 ; x = -B/A

Ax^2 + Bx + 0 = 0 ; then:
x = [-B +- sqrt(B^2 - 4*A*0)] / (2A)
x = [-B +- sqrt(B^2)] / (2A)
x = [-B +- B] / (2A)

x = (-B + B) / (2A) : so x = 0
or
x = (-B - B) / (2A) : so x = -2B / 2A = -B/A

So clearly the results are the same...so why use the quadratic ?
• Jul 25th 2012, 02:34 AM
RonnieSaha
Okay, I am very grateful for all your help on this matter however I really do not understand pretty much any of it. On the sheet that the teacher gave me the title is to Solve Quadratics by Factorising. Then, the first question is x(x - 7) = 0. If someone could please go through this question explaining every step in simple terms as the teacher only tought us what we had to do in 1 lesson?
• Jul 25th 2012, 04:41 AM
Wilmer
Quote:

Originally Posted by RonnieSaha
Okay, I am very grateful for all your help on this matter however I really do not understand pretty much any of it. On the sheet that the teacher gave me the title is to Solve Quadratics by Factorising. Then, the first question is x(x - 7) = 0. If someone could please go through this question explaining every step in simple terms as the teacher only tought us what we had to do in 1 lesson?

If that's the case, then the equation would have been given this way:
x^2 - 7x = 0
You would then "factorize"; that means take out what's common to both terms, which is one "x":
x(x - 7) = 0
Now, if you multiply what's inside brackets by what's outside, you get:
x*x - 7*x = x^2 - 7x : same as what you started with; now you KNOW your factoring is correct.

With x(x - 7) = 0, you have 2 terms: x and x-7.
For their product to equal 0, AT LEAST one of them must equal 0, right?
So it is a MUST that x=0 OR x-7=0; and if x-7=0, then x = 7, right?

Soooo....answer to x^2 - 7x = 0 is x = 0 or x = -7 ; over and out.

And you PROVE it this way:
if x=0: 0^2 - 7*0 = 0
if x=7: 7^2 - 7*7 = 0 - 0 = 0
Got that?

I'm kinda wondering WHY you're having trouble with something this simple;
probably just a temporary "blank"...I get those now and then!
Try going over this IN FULL; you should be ok IF you do:
Factoring in Algebra
• Jul 25th 2012, 08:14 AM
richard1234
Quote:

Originally Posted by RonnieSaha
Okay, I am very grateful for all your help on this matter however I really do not understand pretty much any of it. On the sheet that the teacher gave me the title is to Solve Quadratics by Factorising. Then, the first question is x(x - 7) = 0. If someone could please go through this question explaining every step in simple terms as the teacher only tought us what we had to do in 1 lesson?

Realize that the LHS in \$\displaystyle x(x-7) = 0\$ is already factored. Then either x = 0 or x-7 = 0, by the zero-product property.

Also, please don't accuse your teacher of only teaching you one lesson on this topic. Once you solve it you will realize that it is indeed a very simple problem and can easily be taught in one 15-minute lesson.
• Jul 25th 2012, 01:14 PM
RonnieSaha
Thank you very much for all your help. I can see how I can solve these and can complete my worksheet over the holidays. Once again thank you for explaining it to me :)
• Jul 25th 2012, 01:17 PM
skeeter