# A few general questions and inquiries

• Jul 15th 2006, 07:19 AM
NineZeroFive
A few general questions and inquiries
Hey everyone,

I just joined and have a few math questions and "inquires" about certain things regarding math!

I had a quiz recently . The first question was regarding circles:
$\displaystyle x^2+64$, find the radius (I'm pretty sure it was radius, if not, please feel free to correct me). Can someone explain this?

Is the product of two binomials always a trinomial? Explain please.

Can someone give me some links/information to the following math related subjects:
• Completing the Square
• Equation of a Circle
• Difference of Squares
• Dividing, multiplying exponents
• Solving negative exponents

I'll add more 'topics' as they come to mind.

Thanks for the help, greatly appreciated.
-$\displaystyle NineZeroFive$
• Jul 15th 2006, 08:37 AM
Quick
Completing the Square
Say you are trying to solve the equation $\displaystyle x^2+4x-12=y$ and you are trying to solve for x when y=0

$\displaystyle x^2+4x-12=0\quad\Rightarrow\quad x^2+4x=12$
alright, now you have an equation, all you need to do is factor, the easiest way to do that would be by adding a number such that the left hand side becomes a perfect square, this number is $\displaystyle \left(\frac{b}{2}\right)^2$ b in this case is 4 (it's the coefficient of x) so solve...(remember to add to both sides)

$\displaystyle \left(\frac{b}{2}\right)^2=\left(\frac{4}{2}\right )^2=4$

$\displaystyle x^2+4x=12$

$\displaystyle x^2+4x+4=12+4$

$\displaystyle (x+2)^2=16$

$\displaystyle \sqrt{(x+2)^2}=\sqrt{16}$

$\displaystyle x+2=\pm4$

$\displaystyle x+2=4\quad x+2=-4$

$\displaystyle x=2\quad x=-6$

if you would like an explanation on why this works, please tell me.

~ $\displaystyle Q\!u\!i\!c\!k$
• Jul 15th 2006, 09:12 AM
NineZeroFive
Quote:

Originally Posted by Quick
Say you are trying to solve the equation $\displaystyle x^2+4x-12=y$ and you are trying to solve for x when y=0

$\displaystyle x^2+4x-12=0\quad\Rightarrow\quad x^2+4x=12$
alright, now you have an equation, all you need to do is factor, the easiest way to do that would be by adding a number such that the left hand side becomes a perfect square, this number is $\displaystyle \left(\frac{b}{2}\right)^2$ b in this case is 4 (it's the coefficient of x) so solve...(remember to add to both sides)

$\displaystyle \left(\frac{b}{2}\right)^2=\left(\frac{4}{2}\right )^2=4$

$\displaystyle x^2+4x=12$

$\displaystyle x^2+4x+4=12+4$

$\displaystyle (x+2)^2=16$

$\displaystyle \sqrt{(x+2)^2}=\sqrt{16}$

$\displaystyle x+2=\pm4$

$\displaystyle x+2=4\quad x+2=-4$

$\displaystyle x=2\quad x=-6$

if you would like an explanation on why this works, please tell me.

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

Thanks for the help, an explanation would be great.

Also, would anyone know a formula relating to Quadratic Functions or Geometry (Circles) which uses $\displaystyle \pm$, I learned it a few months ago, just forgot what it was all about. :confused:

-$\displaystyle NineZeroFive$
• Jul 15th 2006, 10:26 AM
Quick
Quote:

Originally Posted by 905
Thanks for the help, an explanation would be great.

$\displaystyle (x+d)^2=(x+d)(x+d)=x^2+dx+dx+d^2=x^2+2dx+d^2$
the equation $\displaystyle x^2+2dx+d^2$ is usually written as $\displaystyle ax^2+bx+c$

now to find $\displaystyle d$ we could divide $\displaystyle b$ by 2, $\displaystyle \left(\frac{b}{2}\right)$

so now we know what $\displaystyle d$ equals. to find $\displaystyle c$ we need to square $\displaystyle d$, so $\displaystyle c=\left(\frac{b}{2}\right)^2$

now remember when we added $\displaystyle c=\left(\frac{b}{2}\right)^2$ to both sides in my last post? you'll notice that the left hand side didn't have $\displaystyle c$ so we added it on.

Quote:

Originally Posted by 905
Also, would anyone know a formula relating to Quadratic Functions or Geometry (Circles) which uses $\displaystyle \pm$, I learned it a few months ago, just forgot what it was all about. :confused:

-$\displaystyle NineZeroFive$

There is a "quadratic formula" $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ that you can use to solve quadratic functions.

FYI the quadratic formula uses the "completing the square" method, watch:

your given equation: $\displaystyle ax^2+bx+c=0$

completing the square only works when there is no coefficient of x^2, so we divide by $\displaystyle a$: $\displaystyle x^2+\frac{b}{a}x+\frac{c}{a}=0$

subtract $\displaystyle \frac{c}{a}$ from both sides: $\displaystyle x^2+\frac{b}{a}x=-\frac{c}{a}$

complete the square: $\displaystyle x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2=\left (\frac{b}{2a}\right)^2-\frac{c}{a}$

factor: $\displaystyle \left(x+\frac{b}{2a}\right)^2=\left(\frac{b}{2a} \right)^2-\frac{c}{a}$

simplify: $\displaystyle \left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}$

combine fractions: $\displaystyle \left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$

find the square root of both sides: $\displaystyle x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}$

simplify: $\displaystyle x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}$

subtract $\displaystyle \frac{b}{2a}$ from both sides: $\displaystyle x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$

compine fractions: $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

~ $\displaystyle Q\!u\!i\!c\!k$
• Jul 15th 2006, 12:07 PM
NineZeroFive
Thanks, exactly what I was looking for.

A few more things:

$\displaystyle ax^2+bx+c=0$ To solve for $\displaystyle c$, doesn't it have to be [/tex] on the left side? Whats the equation $\displaystyle ax^2+bx=c$ then? I've seen it around, don't really know what it means.

Heres another question I remember getting:
$\displaystyle x^2+25$ How many solutions? Is it a complete square? I'm guessing yes, since it x^2=x and 25=5. Correct?

Also, would you know the answer to this: $\displaystyle x^2+64$ What is the radius?
• Jul 15th 2006, 12:53 PM
Quick
I can't answer your questions completely until you tell me what the expressions equal.

Quote:

Originally Posted by 905
Heres another question I remember getting:
$\displaystyle x^2+25$ How many solutions? Is it a complete square? I'm guessing yes, since it x^2=x and 25=5. Correct?

No it's not, always remember the rule $\displaystyle \sqrt{a+b}\neq\sqrt{a}+\sqrt{b}$

Quote:

Originally Posted by 905
Also, would you know the answer to this: $\displaystyle x^2+64$ What is the radius?

as you can see below, this is NOT a circle.
however, $\displaystyle y^2+x^2=64$ is.
• Jul 15th 2006, 12:57 PM
NineZeroFive
Thanks, my memory's probably flawed. :o

You said $\displaystyle y=x^2+5$ is not a complete square. Can you explain why that is and how many solutions it has?

What is the equation $\displaystyle ax^2+bx=c$? Its the same thing as $\displaystyle ax^2+bx+c=0$ except we're solving for $\displaystyle c$ this time correct? Shouldn't $\displaystyle c$ be negative since its brought over to the other side?

Thanks,
-NineZeroFive
• Jul 15th 2006, 01:22 PM
NineZeroFive
Quote:

Originally Posted by Quick
No it's not, always remember the rule $\displaystyle \sqrt{a+b}\neq\sqrt{a}+\sqrt{b}$

Yes, I understand this, but:

$\displaystyle x^2+64$ How do I find out it it is a perfect square? Don't you square both terms and to find out if it is a perfect square? :confused:
• Jul 15th 2006, 01:33 PM
Quick
You are an inquisitive lil' bugger

Quote:

Originally Posted by 905
Thanks, my memory's probably flawed. :o

You said $\displaystyle y=x^2+5$ is not a complete square. Can you explain why that is and how many solutions it has?

there are no solutions, a good way to tell solutions is by using the discriminant of the equation $\displaystyle b^2-4ac$ if the answer is negative, there are no solutions, positive, two solutions, zero yields one solution. Let's try it...
$\displaystyle b^2-4ac=0^2-4(1)(5)=0-20=-20$
there are no solutions to your equation

Quote:

What is the equation $\displaystyle ax^2+bx=c$? Its the same thing as $\displaystyle ax^2+bx+c=0$ except we're solving for $\displaystyle c$ this time correct? Shouldn't $\displaystyle c$ be negative since its brought over to the other side?

Thanks,
-NineZeroFive
I have never seen the equation $\displaystyle ax^2+bx=c$

P.S. I've enclosed an excel program I made for school, only type in boxes with a tab in them (if you hold your mouse over the box than a message will say that you can type in it)
• Jul 15th 2006, 01:39 PM
NineZeroFive
Quote:

Originally Posted by Quick
You are an inquisitive lil' bugger

there are no solutions, a good way to tell solutions is by using the discriminant of the equation $\displaystyle b^2-4ac$ if the answer is negative, there are no solutions, positive, two solutions, zero yields one solution. Let's try it...
$\displaystyle b^2-4ac=0^2-4(1)(5)=0-20=-20$
there are no solutions to your equation

I have never seen the equation $\displaystyle ax^2+bx=c$

P.S. I've enclosed an excel program I made for school, only type in boxes with a tab in them (if you hold your mouse over the box than a message will say that you can type in it)

I'm still a bit confused. You said their were two solutions, then you said there were none. :confused:

Hmmm..

EDIT: Nevermind, I think I understand the equation as well. Since $\displaystyle c$ is not known (whether it is positive or negative)as it a constant, it remains "postive". In a regular situation if it was positive on the right side, it's sign would change before going to the other side.

I think I understand now,

Thanks again,
-NineZeroFive
• Jul 15th 2006, 01:53 PM
Quick
Quote:

Originally Posted by NineZeroFive
EDIT: Nevermind, I think I understand the equation as well. Since $\displaystyle c$ is not known (whether it is positive or negative)as it a constant, it remains "postive". In a regular situation if it was positive on the right side, it's sign would change before going to the other side.

I think I understand now,

Thanks again,
-NineZeroFive

I actually edited that wikipedia entry just to get rid of your confusion. :D

~ $\displaystyle Q\!u\!i\!c\!k$
• Jul 15th 2006, 02:04 PM
Quick
Rules of exponents
$\displaystyle x^a\times x^b=x^{(a+b)}$

$\displaystyle x^a\div x^b=x^{(a-b)}$

$\displaystyle (x^a)^b=x^{(a\times b)}$

$\displaystyle x^{-a}=\frac{1}{x^a}$

If you want an explanation of why these rules work, don't hesitate to ask.

~ $\displaystyle Q\!u\!i\!c\!k$
• Jul 15th 2006, 02:49 PM
NineZeroFive
Quote:

Originally Posted by Quick
$\displaystyle x^a\times x^b=x^{(a+b)}$

$\displaystyle x^a\div x^b=x^{(a-b)}$

$\displaystyle (x^a)^b=x^{(a\times b)}$

$\displaystyle x^{-a}=\frac{1}{x^a}$

If you want an explanation of why these rules work, don't hesitate to ask.

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

Sure, why not. I know a few of those already, just dont know why.

I've learn't alot today,

Thanks again.
-NineZeroFive
• Jul 15th 2006, 04:36 PM
Quick
Quote:

Originally Posted by NineZeroFive
Sure, why not. I know a few of those already, just dont know why.

I've learn't alot today,

Thanks again.
-NineZeroFive

Apparently I'm a good teacher :cool:

anyway,

$\displaystyle x^a\times x^b=x^{(a+b)}$
let's do a sample equation...

$\displaystyle a^3\times a^2=a^5$

write it out: $\displaystyle \underbrace{\overbrace{a\times a\times a}^{a^3}\times \overbrace{a\times a}^{a^2}}_{a^5}$

next rule:
$\displaystyle x^a\div x^b=x^{(a-b)}$

back to our sample equation:
$\displaystyle a^3\div a^2=a^1$

write it out:$\displaystyle \frac{a^3}{a^2}=\frac{a\times a\times a}{a\times a}=\frac{\not a\times \not a\times a}{\not a\times \not a}=a^1$

next rule: $\displaystyle (x^a)^b=x^{(a\times b)}$

our sample equation:
$\displaystyle (a^3)^2=a^6$

write it out:$\displaystyle (a^3)^2=a^3\times a^3=\underbrace{\overbrace{a\times a\times a}^{a^3}\times \overbrace{a\times a\times a}^{a^3}}_{a^6}$

next rule: $\displaystyle x^{-a}=\frac{1}{x^a}$

this one is not so easy to explain, but here I go...

apply rule #2 to our sample equation:
$\displaystyle a^2\div a^3=a^{(2-3)}=a^{-1}$

write it out: $\displaystyle \frac{a\times a}{a\times a\times a}=\frac{\not a\times \not a}{\not a\times \not a\times a}=\frac{1}{a}$

that is why $\displaystyle a^{-1}=\frac{1}{a}$

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