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Math Help - Quadratic Formulas!

  1. #1
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    Quadratic Formulas!

    I know I have to use the equation:

    -b (plus or negative)(sqrt)b-4ac
    x= ---------------------------------
    2a

    The question I have is x-2x-5=10. I have ot use the completed square form to solve the equation but im confused.


    Also another question I have is to investigate by substituting values for x satisfying -5(Larger than or equal to) x (Lessthan or equal to) 5, find the largest possible value of:

    6
    --------
    x-2x-5

    And then justify why my chosen value of x maximises the equation.


    Sorry for the long questions and stuff but if oyu wouldnt mind explaining how you did them that would help me pestering you next time
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Yppolitia View Post
    The question I have is x-2x-5=10. I have ot use the completed square form to solve the equation but im confused.
    x^2 - 2x - 5 = 0

    x^2 - 2x = 5

    Now, we want to add a number to both sides such that the LHS will be a perfect square. As it happens x^2 + 2a + a^2 = (x + a)^2. Comparing the coefficient of the linear term of this form with the linear term in the equation above we can see that
    2a = -2
    a = -1

    Thus a^2 = (-1)^2 = 1. So add 1 to both sides:
    x^2 - 2x + 1 = 5 + 1
    (x - 1)^2 = 6
    x - 1 = (+/-)sqrt(6)
    x = 1 (+/-) sqrt(6)

    We can check this using the quadratic formula:
    x^2 - 2x - 5 = 0
    Gives
    x = {-(-2) (+/-) sqrt([-2]^2 - 4*1*(-5))}/(2*1)

    x = {2 (+/-) sqrt(24)}/2 = {2 (+/-) sqrt(4*6)}/2 = {2 (+/-) 2*sqrt(6)}/2

    x = 1 (+/-) sqrt(6) (Check!)

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Yppolitia View Post
    Also another question I have is to investigate by substituting values for x satisfying -5(Larger than or equal to) x (Lessthan or equal to) 5, find the largest possible value of:

    6
    --------
    x-2x-5

    And then justify why my chosen value of x maximises the equation.
    I'm not sure I understand the question. According to the first problem the denominator of this function goes to zero for x = 1 (+/-) sqrt(6) indicating we have vertical asymptotes at these x values. So the "largest possible value" of the function is positive infinity, which really means there is no maximum.

    -Dan
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  4. #4
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    Quote Originally Posted by topsquark View Post
    x^2 - 2x - 5 = 0

    x^2 - 2x = 5

    Now, we want to add a number to both sides such that the LHS will be a perfect square. As it happens x^2 + 2a + a^2 = (x + a)^2. Comparing the coefficient of the linear term of this form with the linear term in the equation above we can see that
    2a = -2
    a = -1

    Thus a^2 = (-1)^2 = 1. So add 1 to both sides:
    x^2 - 2x + 1 = 5 + 1
    (x - 1)^2 = 6
    x - 1 = (+/-)sqrt(6)
    x = 1 (+/-) sqrt(6)

    We can check this using the quadratic formula:
    x^2 - 2x - 5 = 0
    Gives
    x = {-(-2) (+/-) sqrt([-2]^2 - 4*1*(-5))}/(2*1)

    x = {2 (+/-) sqrt(24)}/2 = {2 (+/-) sqrt(4*6)}/2 = {2 (+/-) 2*sqrt(6)}/2

    x = 1 (+/-) sqrt(6) (Check!)

    -Dan


    I dont get how you've gone from x²-2x-5=10 to x²-2x-5=0
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  5. #5
    Member Glaysher's Avatar
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    Quote Originally Posted by Yppolitia View Post
    I dont get how you've gone from x²-2x-5=10 to x²-2x-5=0
    Misread I assume

    x² - 2x - 5 = 10
    x² - 2x - 15 = 0

    Factorise

    (x - 5)(x + 3) = 0

    Solutions

    x = 5 and x = -3

    EDIT
    Oh complete the square

    (x - 1)² - 1 - 15 = 0
    (x - 1)² - 16 = 0
    (x - 1)² = 16
    x - 1 = + 4 or x - 1 = - 4
    x = 5 or x = -3
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  6. #6
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    Quote Originally Posted by Yppolitia View Post
    I know I have to use the equation:

    -b (plus or negative)(sqrt)b-4ac
    x= ---------------------------------
    2a

    The question I have is x-2x-5=10. I have ot use the completed square form to solve the equation but im confused.
    Hi,

    x-2x = 15

    complete the square on the LHS by adding 1. What you do on the LHS you have to do on the RHS too!

    x+2x+1 = 15+1

    (x-1) = 4

    Calculate the square-root on both sides:

    x - 1 = 4 or x - 1 = -4

    So x = 1 + 4 = 5 or x = 1 - 4 = -3

    Quote Originally Posted by Yppolitia View Post
    Also another question I have is to investigate by substituting values for x satisfying -5(Larger than or equal to) x (Lessthan or equal to) 5, find the largest possible value of:
    6
    --------
    x-2x-5

    And then justify why my chosen value of x maximises the equation....
    As topsquark has already shown this term is undefined for x = 1 +/- sqrt(6). I don't know if you know to calculate a maximum or a minimum by using derivatives. If not, you can use another method:
    A fraction which numerator is a constant will be extremly great if the denominator is extremely small. The denominator is a quadratic term. The smallest value correspond to the vertex of a parabola.
    x-2x-5 = x-2x+1-1-5 = (x-1) - 6.

    So at x = 1 the smallest value of the quadratic term is -6.

    So your fraction has a maximum at x = 1 and the maximum has the value 6/(-6) = -1
    I've attached a graph to demonstrate what I've calculated.
    Attached Thumbnails Attached Thumbnails Quadratic Formulas!-gebr_fkt051006.gif  
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  7. #7
    Member Glaysher's Avatar
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    Quote Originally Posted by topsquark View Post
    I'm not sure I understand the question. According to the first problem the denominator of this function goes to zero for x = 1 (+/-) sqrt(6) indicating we have vertical asymptotes at these x values. So the "largest possible value" of the function is positive infinity, which really means there is no maximum.

    -Dan
    I suspect a mistype in the question as if the minimum point of the function in the denominator was above the x axis then this point would correspond to the maximum of the whole function
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Yppolitia View Post
    I dont get how you've gone from x-2x-5=10 to x-2x-5=0
    Sorry. Yes, I misread the question. But if you can follow the method that, at least, is correct. Earboth has it done with the correction.

    -Dan
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