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Math Help - calculation to an equation?

  1. #1
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    calculation to an equation?

    4*(5x+3)^2=9

    I am havning problems with the process to solve this Equation.
    This is how far i have come:

    4*(5x+3)^2=9
    4*(5x^2+30x+9)=9 Using a rule here that i dont know the name of in english
    20x^2+120x+36=9
    20x^2+120x=-27
    20x^2+x=-27/120 now what? How do I get rid of the "^2" from 20x? or is this totaly wrong?

    Fast answers are very much appreciated here!

    Thank you in advance!

    Rickard Liljeros
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by liljeros View Post
    4*(5x+3)^2=9

    I am havning problems with the process to solve this Equation.
    This is how far i have come:

    4*(5x+3)^2=9
    4*(5x^2+30x+9)=9 Using a rule here that i dont know the name of in english << SiMoon says : "expanding" or "developping" or the quadratics formula ?
    20x^2+120x+36=9
    20x^2+120x=-27
    20x^2+x=-27/120 now what? How do I get rid of the "^2" from 20x? or is this totaly wrong?

    Fast answers are very much appreciated here!

    Thank you in advance!

    Rickard Liljeros
    Don't bother

    4=2^2

    \implies 4*(5x+3)^2=\left(2(5x+3)\right)^2=\boxed{(10x+6)^2  }

    Therefore :
    4*(5x+3)^2=9 \Longleftrightarrow (10x+6)^2=9 \Longleftrightarrow (10x+6)^2-3^2=0

    Now, remember that a^2-b^2=(a-b)(a+b)

    And it's all done
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  3. #3
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    Thank you very much! That helped me alot, I need to learn to try out more then one way in the begining
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  4. #4
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    Quote Originally Posted by Moo View Post
    Hello,



    Don't bother

    4=2^2

    \implies 4*(5x+3)^2=\left(2(5x+3)\right)^2=\boxed{(10x+6)^2  }

    Therefore :
    4*(5x+3)^2=9 \Longleftrightarrow (10x+6)^2=9 \Longleftrightarrow (10x+6)^2-3^2=0

    Now, remember that a^2-b^2=(a-b)(a+b)

    And it's all done
    Now I fond one thing here i didnt understand, when u multiply the "4" in to the "()" you dont get 10 and 6? isnt 5*4=20 and 4*3=12
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  5. #5
    Moo
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    Quote Originally Posted by liljeros View Post
    Now I fond one thing here i didnt understand, when u multiply the "4" in to the "()" you dont get 10 and 6? isnt 5*4=20 and 4*3=12
    No, it's 4*(5x+3)^2=2^2*(5x+3)^2

    In general, (ab)^n=a^n * b^n

    So here, it's equal to (2*(5x+3))^2

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  6. #6
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    Quote Originally Posted by Moo View Post
    No, it's 4*(5x+3)^2=2^2*(5x+3)^2

    In general, (ab)^n=a^n * b^n

    So here, it's equal to (2*(5x+3))^2

    Ahaa! You're a smart one! Thank you very much!
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  7. #7
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    Try this:

    4(5x+3)^2 = 9

    1. Divide both sides by 4

    (5x+3)^2 = \frac{9}{4}

    2. Take square root of both sides

    5x + 3 = \pm\frac{3}{2}

    3. Solve

    5x + 3 = \frac{3}{2} and 5x + 3 = \frac{-3}{2}

    x = \frac{-3}{10} and x = \frac{-9}{10}
    Last edited by masters; May 20th 2008 at 01:45 PM.
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  8. #8
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    Quote Originally Posted by masters View Post
    Try this:

    4(5x+3)^2 = 9

    1. Divide both sides by 4

    (5x+3)^2 = \frac{9}{4}

    2. Take square root of both sides

    5x + 3 = \pm\frac{3}{2}

    3. Solve

    5x + 3 = \frac{3}{2} and 5x + 3 = \frac{-3}{2}

    x = \frac{-3}{10} and x = \frac{-9}{10}

    That was one other good way to solve it, maybe a bit simpler... Wonder wich I should show my teacher. Well i could just show them both

    Thank you very much for the help!
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  9. #9
    A riddle wrapped in an enigma
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    Go for it. Be a rebel.
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  10. #10
    Moo
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    Quote Originally Posted by liljeros View Post
    That was one other good way to solve it, maybe a bit simpler... Wonder wich I should show my teacher. Well i could just show them both

    Thank you very much for the help!
    It depends on what you're studying now
    Or make a survey in your class to see which one is the most original one
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