Some more general questions..

**NineZeroFive** · July 18th 2006, 11:53 AM

Hey,

I have some general questions from "all-around" math:

When factoring, why do you have to use perfect squares?
Is the equation ax^2+bx+c=0 the same as the equation y=ax^2+bx+c? (referring to quadratics)
What does the term "difference of squares" imply?
Are there any general rules when factoring trinomials?

Thanks everyone.
-NineZeroFive

**ThePerfectHacker** · July 18th 2006, 11:59 AM

This,
$ax^2+bx+c=0$ is an equation. Meaning it has to be solved.

This,
$ax^2+bx+c=y$ is a function. It represents the relationship between the x-coordinate and y-coordinate.
---
Diffrence of square means,
$x^2-y^2$ and they can always be factored.
$(x+y)(x-y)$
---
Why use perfect squares?
You mean why bring everything to perfect square in a conic? Because after you used perfect squares the shape of the conic is easily recognizable. And also it can be graphed easily by simply doing a translation of the axes.

**NineZeroFive** · July 18th 2006, 12:09 PM

Originally Posted by ThePerfectHacker

This,
$ax^2+bx+c=0$ is an equation. Meaning it has to be solved.

This,
$ax^2+bx+c=y$ is a function. It represents the relationship between the x-coordinate and y-coordinate.
---
Diffrence of square means,
$x^2-y^2$ and they can always be factored.
$(x+y)(x-y)$
---
Why use perfect squares?
You mean why bring everything to perfect square in a conic? Because after you used perfect squares the shape of the conic is easily recognizable. And also it can be graphed easily by simply doing a translation of the axes.

Thanks for the FAST reply!

I mean..why must we use perfect squares when factoring? :S

**NineZeroFive** · July 18th 2006, 12:17 PM

Also, what is the x intercept(s) and y intercept of $y=x^2$ ? I'm pretty sure I have the correct answer, just want to confirm because the answer on the back of the textbook I'm using seems 'wierd'. It says the x-intercept is 12 (??) and y inctercept is 0 (which I understand).

Can someone please use the Quadractic formula to figure out the x-intercept(s) of the above equation, if there are any?

Thanks again,
-NineZeroFive

**Jameson** · July 18th 2006, 12:52 PM

The x-intercept would be where the graph crosses/touches the x-axis. Thus at this point y=0. So this occurs when 0=x^2, obviously at 0.

The y-intercept would be where the graph crosses/touches the y-axis. Thus at this point x=0. So this occurs when y=0^2, or 0.

**NineZeroFive** · July 18th 2006, 12:56 PM

Originally Posted by Jameson

The x-intercept would be where the graph crosses/touches the x-axis. Thus at this point y=0. So this occurs when 0=x^2, obviously at 0.

The y-intercept would be where the graph crosses/touches the y-axis. Thus at this point x=0. So this occurs when y=0^2, or 0.

Thats, what I thought, Thanks, I guess I'll have to ask the teacher tomorrow.

Three more questions: (both referring to parabolas)

If an expression is in the form of $ax^2+bx+c$ , and the question asks you to find the y-intercept, how would you do so?

Also, when using the Quadratic formula, how to you know when there is only 1 x-intercept instead of the usual 2?

Can someone please use the Quadractic formula to figure out the x-intercept(s) of the equation $y=x^2$ , if there are any?

Thanks alot.
-NineZeroFive

**Quick** · July 18th 2006, 04:18 PM

Originally Posted by NineZeroFive

Thats, what I thought, Thanks, I guess I'll have to ask the teacher tomorrow.

Three more questions: (both referring to parabolas)

If an expression is in the form of $ax^2+bx+c$ , and the question asks you to find the y-intercept, how would you do so?

solve for x=0

Also, when using the Quadratic formula, how to you know when there is only 1 x-intercept instead of the usual 2?

there is a thing called the discriminant $b^2-4ac$ (it is the thing under the radical in the quadratic formula) if the discriminant is negative there is no x-intercept, if positive there are 2 x-intercepts, and if it equals 0 than there is 1 x-intercept.

$x=\frac{\neg b\pm\sqrt{\bold{b^2-4ac}}}{2a}$
(the discriminant is bold)

Can someone please use the Quadractic formula to figure out the x-intercept(s) of the equation $y=x^2$ , if there are any?

Thanks alot.
-NineZeroFive

write out the equation in full...
$y=ax^2+bx+c\quad\Rightarrow\quad 0=1x^2+0x+0$

than solve...

$x=\frac{\neg b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{\neg 0\pm\sqrt{0^2-4(1)(0)}}{2(1)}$

$x=\frac{\pm\sqrt{0}}{2}$ the discriminant is 0, therefore we know there is only 1 solution.

~ $Q\!u\!i\!c\!k$

**NineZeroFive** · July 18th 2006, 04:20 PM

Originally Posted by Quick

solve for x=0
there is a thing called the discriminant $b^2-4ac$ (it is the thing under the radical in the quadratic formula) if the discriminant is negative there is no x-intercept, if positive there are 2 x-intercepts, and if it equals 0 than there is 1 x-intercept.

$x=\frac{\neg b\pm\sqrt{\bold{b^2-4ac}}}{2a}$
(the discriminant is bold)

write out the equation in full...
$y=ax^2+bx+c\quad\Rightarrow\quad 0=1x^2+0x+0$

than solve...

$x=\frac{\neg b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{\neg 0\pm\sqrt{0^2-4(1)(0)}}{2(1)}$

$x=\frac{\pm\sqrt{0}}{2}$ the discriminant is 0, therefore we know there is only 1 solution.

~ $Q\!u\!i\!c\!k$

Thanks ALOT Quick. YOU ARE AMAZING!!

BTW: Your PM inbox is full.

**Quick** · July 18th 2006, 04:27 PM

Originally Posted by NineZeroFive

Three more questions: (both referring to parabolas)

This is one of the funniest mistakes in grammar I've ever seen.

Also, I wanted to say that when factoring you don't need to make things into perfect squares, it's just a convenient thing to do.

**Quick** · July 18th 2006, 04:33 PM

Originally Posted by NineZeroFive

Hey,

I have some general questions from "all-around" math:

Are there any general rules when factoring trinomials?

Thanks everyone.
-NineZeroFive

Yes, but they're merely "tricks" to help you factor.

when given an expression in the form $x^2+bx+c$ the factors will be in the form $(x+d)(x+g)$

when given an expression in the form $x^2-bx+c$ the factors will be in the form $(x-d)(x-g)$

when given an expression in the form $x^2+bx-c$ the factors will be in the form $(x+d)(x-g)$ where $d>g$

when given an expression in the form $x^2-bx-c$ the factors will be in the form $(x+d)(x-g)$ where $d<g$

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