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Thread: A few general questions and inquiries

  1. #1
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    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.
    -$\displaystyle NineZeroFive$
    Last edited by NineZeroFive; Jul 15th 2006 at 08:51 AM.
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  2. #2
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    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.

    ~ $\displaystyle Q\!u\!i\!c\!k$
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  3. #3
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    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.

    -$\displaystyle NineZeroFive$
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  4. #4
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    Quadratic Formula

    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.

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


    ~ $\displaystyle Q\!u\!i\!c\!k$
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  5. #5
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    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?
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  6. #6
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    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.
    Attached Thumbnails Attached Thumbnails A few general questions and inquiries-unknown-equation.jpg  
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  7. #7
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    Thanks, my memory's probably flawed.

    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
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  8. #8
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    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?
    Last edited by NineZeroFive; Jul 15th 2006 at 01:36 PM.
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  9. #9
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    You are an inquisitive lil' bugger

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

    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


    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)
    Attached Files Attached Files
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  10. #10
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    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)
    Link to equation: http://en.wikipedia.org/wiki/Completing_the_square

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

    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
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  11. #11
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    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.

    ~ $\displaystyle Q\!u\!i\!c\!k$
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  12. #12
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    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.

    ~ $\displaystyle Q\!u\!i\!c\!k$
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  13. #13
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    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
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  14. #14
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    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

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

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