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Math Help - solutions of quadratic equations

  1. #1
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    solutions of quadratic equations

    Hi, i am struggling with factorising the quadratic equations. I have got to grips with basic factorisation. any help?

    1. solve the following quadratic equations by factorisation:

    a)
    (x+6)(x-3)=22
    ......
    i have tried this one and came out with this answer:
    x(x-3)+6(x-3)=22
    x^2-3x+6x-18=22
    x^2+3x-22-16=0
    x^2+3x-38
    (x+19)(x-2)

    Is this correct?

    b) 20x=4+3/x

    c) 4t(3t-2)=3-8t


    2.
    a) x^2=7x+4

    b) (3x+4)(x-1)=1

    c) 1/2x-7=x/x-1
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  2. #2
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    Quote Originally Posted by daniel123 View Post
    Hi, i am struggling with factorising the quadratic equations. I have got to grips with basic factorisation. any help?

    1. solve the following quadratic equations by factorisation:

    a)
    (x+6)(x-3)=22

    x^2 + 3x - 18 = 22

    x^2 + 3x - 40 = 0

    (x + 8)(x - 5) = 0

    set each factor = 0 and solve.


    b) 20x=4+3/x

    multiply every term by x ...

    20x^2 = 4x + 3

    20x^2 - 4x - 3 = 0

    factor, set each factor = 0 and solve


    c) 4t(3t-2)=3-8t

    12t^2 - 8t = 3 - 8t

    set = 0 and solve as before



    2.
    a) x^2=7x+4

    set = 0 and solve as before

    b) (3x+4)(x-1)=1

    expand, set = 0 and solve

    c) 1/(2x-7)=x/(x-1)

    note the parentheses.

    cross multiply ...

    x(2x-7) = 1(x-1)

    distribute, set = 0 and solve


    ...
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  3. #3
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    i have the answers to 1.
    just to check im going in the right direction
    a)factors x=-8 x=5
    b)factors x=3/4 x=-1/5
    c)factors x=-3/3 x=1/4

    I am struggling still with question 2.
    so far i have worked these out as fractions. i need to change them into an answer with 3 decimal places, but i do not know if i have the correct answer....
    a)factors x=4 x=-1
    b)factors x=5/3 x=-1
    c)factors x=-1/2 x=-1
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  4. #4
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    Quote Originally Posted by daniel123 View Post
    2.
    a) x^2=7x+4
    \displaystyle x^2=7x+4

    \displaystyle x^2-7x-4=0

    Completing the square

    \displaystyle x^2-7x+\frac{49}{4}-4-\frac{49}{4}=0

    \displaystyle \left(x-\frac{7}{2}\right)^2-4-\frac{49}{4}=0

    \displaystyle \left(x-\frac{7}{2}\right)^2-\frac{65}{4}=0

    \displaystyle \left(x-\frac{7}{2}\right)^2-\left(\frac{\sqrt{65}}{2}\right)^2=0

    Now apply the difference of 2 squares


    Quote Originally Posted by daniel123 View Post


    b) (3x+4)(x-1)=1
    Expand this one out first and make RHS = 0, then have another attempt.



    Quote Originally Posted by daniel123 View Post


    c) 1/2x-7=x/x-1
    What is this equation?

    \displaystyle \frac{1}{2x-7} = \frac{x}{x-1}

    or

    \displaystyle \frac{1}{2x}-7 = \frac{x}{x}-1

    Use brackets to provide clarity
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  5. #5
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    I have tried an alternate way shown in an exercise book for question 2. I have attemted a) . Just woundering if i was going in the right direction?
    2.
    a)x^2=7x+4
    b)(3x+4)(x-1)=1
    c)1/(2x-7) = x/(x-1)


    2.
    a)
    Roots are x=7.531 x=-0.531
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  6. #6
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    Quote Originally Posted by daniel123 View Post

    2.
    a)
    Roots are x=7.531 x=-0.531
    These look good.



    Quote Originally Posted by daniel123 View Post

    b)(3x+4)(x-1)=1
    Did you expand this one yet?


    Quote Originally Posted by daniel123 View Post

    c)1/(2x-7) = x/(x-1)

    \displaystyle \frac{1}{2x-7} = \frac{x}{x-1}

    Cross multiplying

    \displaystyle x-1 = x(2x-7)

    \displaystyle x-1 = 2x^2-7x

    \displaystyle  2x^2-8x+1=0
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