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Thread: Quadratic equations using the formula method

  1. #1
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    Quadratic equations using the formula method

    Can anyone solve the following quadratic equation using the formula method...

    Xsquared - 10x + 3 = 0

    Please show working out so it is possible for me to follow and see where i am going wrong

    Thanks
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by duckegg911 View Post
    Can anyone solve the following quadratic equation using the formula method...

    Xsquared - 10x + 3 = 0

    Please show working out so it is possible for me to follow and see where i am going wrong

    Thanks
    Let me try again...

    $\displaystyle x^2 - 10x + 3 = 0$

    It seems he's allowed to use the quadratic formula.

    All quadratic equations are written in the format of $\displaystyle ax^2 + bx + c = 0$

    That means that $\displaystyle a$ is the co-efficient of the$\displaystyle x^2$ ; $\displaystyle b$ is the co-efficient of $\displaystyle x$ ; and $\displaystyle c$ is the constant.

    According to the quadratic formula:

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

    All you have to do is substitute the corresponding co-efficients into the formula.
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  3. #3
    Super Member wingless's Avatar
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    A standard quadratic equation is in the form
    $\displaystyle ax^2 + bx + c = 0$
    a, b and c being the coefficients.

    The formula for the discriminant ($\displaystyle \Delta$) is
    $\displaystyle \Delta = b^2 - 4ac$

    If $\displaystyle \Delta > 0$, there are 2 different roots.
    If $\displaystyle \Delta = 0$, there are 2 equivalent roots. ($\displaystyle x_1=x_2$)
    If $\displaystyle \Delta < 0$, there are no real roots.

    The roots are,

    $\displaystyle x_1 = \frac{-b - \sqrt{\Delta}}{2a}$

    $\displaystyle x_2 = \frac{-b + \sqrt{\Delta}}{2a}$

    Now plug a, b and c in these formulas.
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  4. #4
    Junior Member Pinsky's Avatar
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    Let's say we write your equation this way.

    $\displaystyle ax^2+bx+c=0$
    where:
    $\displaystyle a=1$
    $\displaystyle b=-10$
    $\displaystyle c=3$

    now we have the coefficients we use in the equation.

    $\displaystyle x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

    Now you just put a,b and c into the equation and that's it.

    There are usually 2 results for x.
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  5. #5
    Junior Member Pinsky's Avatar
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    Sorry, was writing the answer while you posted them.
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  6. #6
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Pinsky View Post
    Sorry, was writing the answer while you posted them.
    No problem, at least now (s)he has 3 explanations to choose from...
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  7. #7
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    Quote Originally Posted by Pinsky View Post
    Sorry, was writing the answer while you posted them.
    Thanks to you both, ill take a long hard look at these explanations and let you both know how i get on ,

    thanks again
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  8. #8
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    Quote Originally Posted by janvdl View Post
    No problem, at least now (s)he has 3 explanations to choose from...
    Not to mention the explanations accompanying the examples s/he should have in their class notes .....
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  9. #9
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    Quote Originally Posted by janvdl View Post
    Let me try again...

    $\displaystyle x^2 - 10x + 3 = 0$

    It seems he's allowed to use the quadratic formula.

    All quadratic equations are written in the format of $\displaystyle ax^2 + bx + c = 0$

    That means that $\displaystyle a$ is the co-efficient of the$\displaystyle x^2$ ; $\displaystyle b$ is the co-efficient of $\displaystyle x$ ; and $\displaystyle c$ is the constant.

    According to the quadratic formula:

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

    All you have to do is substitute the corresponding co-efficients into the formula.
    Are the answers for -9.6904 and -3.095 ???
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  10. #10
    Bar0n janvdl's Avatar
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    Quote Originally Posted by duckegg911 View Post
    Are the answers for -9.6904 and -3.095 ???
    I get positive answers, not negative. That's the only problem with your answers.
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  11. #11
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    Quote Originally Posted by janvdl View Post
    I get positive answers, not negative. That's the only problem with your answers.
    This is because - plus a - = a + ???
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  12. #12
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    Quote Originally Posted by duckegg911 View Post
    This is because - plus a - = a + ???
    sorry that sound silly....

    its just i cant follow why you get possitive answers...

    the answers you have are 9.6904 and 3.095 ?
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  13. #13
    Bar0n janvdl's Avatar
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    After substitution into the formula, your formula should look like this:

    $\displaystyle x = \frac{ +10 \pm \sqrt{100 - 4(1)(3)} }{2}$

    Does it?
    Last edited by janvdl; Jan 18th 2008 at 05:20 AM. Reason: forgot \sqrt
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  14. #14
    Bar0n janvdl's Avatar
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    Here's the full solution. Ask me about any step you do not understand/

    $\displaystyle x = \frac{ +10 \pm \sqrt{100 - 4(1)(3)} }{2}$

    $\displaystyle x = \frac{ +10 \pm \sqrt{100 - 12} }{2}$

    $\displaystyle x = \frac{ +10 \pm \sqrt{88} }{2}$

    BUT $\displaystyle \sqrt{88} = \sqrt{4} \times \sqrt{22}$

    AND we know $\displaystyle \sqrt{4} = 2$

    $\displaystyle x = \frac{ +10 \pm \sqrt{4} \cdot \sqrt{22} }{2}$

    $\displaystyle x = \frac{ +10 \pm 2 \sqrt{22} }{2}$

    Divide 10 by 2, and divide $\displaystyle 2 \sqrt{22}$ by 2

    $\displaystyle x = +5 \pm \sqrt{22}$
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  15. #15
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    Quote Originally Posted by janvdl View Post
    Here's the full solution. Ask me about any step you do not understand/

    $\displaystyle x = \frac{ +10 \pm \sqrt{100 - 4(1)(3)} }{2}$

    $\displaystyle x = \frac{ +10 \pm \sqrt{100 - 12} }{2}$

    $\displaystyle x = \frac{ +10 \pm \sqrt{88} }{2}$

    BUT $\displaystyle \sqrt{88} = \sqrt{4} \times \sqrt{22}$

    AND we know $\displaystyle \sqrt{4} = 2$

    $\displaystyle x = \frac{ +10 \pm \sqrt{4} \cdot \sqrt{22} }{2}$

    $\displaystyle x = \frac{ +10 \pm 2 \sqrt{22} }{2}$

    Divide 10 by 2, and divide $\displaystyle 2 \sqrt{22}$ by 2

    $\displaystyle x = +5 \pm \sqrt{22}$
    I have -10 all the way through rather than +10 ???
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