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Math Help - Another square technique...

  1. #1
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    Another square technique...

    Use the technique of completing the square to transform the quadratic equation into the form (x + c)2 = a.

    2x2 + 32x + 12 = 0

    Thanks alot guys.
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by tysonrss View Post
    Use the technique of completing the square to transform the quadratic equation into the form (x + c)2 = a.

    2x2 + 32x + 12 = 0

    Thanks alot guys.
    2x^{2} + 32x + 12 = 0

    2(x^{2} + 16x + 6 = 0)

    x^{2} + 16x + 6 = 0

    x^2 + 2\times 8\times x + 6 = 0

    Can you take it further? Please try
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  3. #3
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    Quote Originally Posted by harish21 View Post
    2x^{2} + 32x + 12 = 0

    2(x^{2} + 16x + 6 = 0)

    x^{2} + 16x + 6 = 0

    x^2 + 2\times 8\times x + 6 = 0

    Can you take it further? Please try
    Would it be (x + 6)2 = 58 ?
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by tysonrss View Post
    Would it be (x + 6)2 = 58 ?
    No.

    You are left with x^2+ 2\times x \times 8 + 6 = 0.....(I)

    So (x+c)^2 = x^2+ 2.x.c +c^2...(II)

    if you compare I and II, you should see that c = 8.

    So you need to have (x+8)^2, which is equal to x^2 + 2.x.8+ 64

    now if you look at equation (I) again, you already have x^2+ 2\times x \times 8 + 6 = 0.

    To make this x^2+ 2\times x \times 8 + 64 = 0,

    you need to add 58 right?

    so,

    x^2+ 2\times x \times 8 + 6 +58 -58= 0

    Now put the above equation in the form of (x+c)^2=a
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  5. #5
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    Quote Originally Posted by harish21 View Post
    No.

    You are left with x^2+ 2\times x \times 8 + 6 = 0.....(I)

    So (x+c)^2 = x^2+ 2.x.c +c^2...(II)

    if you compare I and II, you should see that c = 8.

    So you need to have (x+8)^2, which is equal to x^2 + 2.x.8+ 64

    now if you look at equation (I) again, you already have x^2+ 2\times x \times 8 + 6 = 0.

    To make this x^2+ 2\times x \times 8 + 64 = 0,

    you need to add 58 right?

    so,

    x^2+ 2\times x \times 8 + 6 +58 -58= 0

    Now put the above equation in the form of (x+c)^2=a
    (<i>x</i> + 8)2 = 58 would be it than right?
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  6. #6
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by tysonrss View Post
    (<i>x</i> + 8)2 = 58 would be it than right?
    yes
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