Page 1 of 2 12 LastLast
Results 1 to 15 of 19

Math Help - Quadratic equations using the formula method

  1. #1
    Junior Member
    Joined
    Jan 2008
    Posts
    31

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    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...

    x^2 - 10x + 3 = 0

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

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

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

    According to the quadratic formula:

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

    All you have to do is substitute the corresponding co-efficients into the formula.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    A standard quadratic equation is in the form
    ax^2 + bx + c = 0
    a, b and c being the coefficients.

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

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

    The roots are,

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

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

    Now plug a, b and c in these formulas.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member Pinsky's Avatar
    Joined
    Jan 2008
    From
    Rijeka
    Posts
    44
    Let's say we write your equation this way.

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

    now we have the coefficients we use in the equation.

    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member Pinsky's Avatar
    Joined
    Jan 2008
    From
    Rijeka
    Posts
    44
    Sorry, was writing the answer while you posted them.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    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...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jan 2008
    Posts
    31
    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
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    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 .....
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jan 2008
    Posts
    31
    Quote Originally Posted by janvdl View Post
    Let me try again...

    x^2 - 10x + 3 = 0

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

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

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

    According to the quadratic formula:

    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 ???
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    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.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Jan 2008
    Posts
    31
    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 + ???
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Jan 2008
    Posts
    31
    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 ?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    After substitution into the formula, your formula should look like this:

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

    Does it?
    Last edited by janvdl; January 18th 2008 at 05:20 AM. Reason: forgot \sqrt
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Here's the full solution. Ask me about any step you do not understand/

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

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

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

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

    AND we know \sqrt{4} = 2

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

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

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

    x = +5 \pm \sqrt{22}
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Junior Member
    Joined
    Jan 2008
    Posts
    31
    Quote Originally Posted by janvdl View Post
    Here's the full solution. Ask me about any step you do not understand/

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

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

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

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

    AND we know \sqrt{4} = 2

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

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

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

    x = +5 \pm \sqrt{22}
    I have -10 all the way through rather than +10 ???
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Quadratic formula help
    Posted in the Algebra Forum
    Replies: 6
    Last Post: March 28th 2010, 07:44 PM
  2. Quadratic Equations Using a Formula
    Posted in the Algebra Forum
    Replies: 11
    Last Post: January 2nd 2010, 05:30 AM
  3. Quadratic formula 2
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 28th 2008, 02:15 PM
  4. Replies: 1
    Last Post: June 12th 2008, 09:30 PM
  5. Replies: 2
    Last Post: September 18th 2007, 05:20 PM

Search Tags


/mathhelpforum @mathhelpforum