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Math Help - Quadratic equation - help

  1. #1
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    Quadratic equation - help

    Code:
    h=16t^2+vt+s (Solve for t)
    h-h=16t^2+vt+s-h
    0=16t^2+vt+s-h
    a=16,b=v,c=s-h
    t= (-b[sqrt](b^2-4ac))/2a
    t=(-v[sqrt]((v)^2-4(16)(s-h) ))/2(16) 
    t=(-v[sqrt](v^2-4(16s-16h) ))/32
    t=(-v[sqrt](v^2-64s+64h))/32
    t=(-vv+8s+8h)/32
    t=(-v)/32 v/32+8s/32+8h/32
    Is this correct??
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  2. #2
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    Quote Originally Posted by mcruz65 View Post
    h=16t^2+vt+s (Solve for t)
    h-h=16t^2+vt+s-h
    0=16t^2+vt+s-h
    a=16,b=v,c=s-h
    t= (-b[sqrt](b^2-4ac))/2a
    t=(-v[sqrt]((v)^2-4(16)(s-h) ))/2(16)
    t=(-v[sqrt](v^2-4(16s-16h) ))/32
    t=(-v[sqrt](v^2-64s+64h))/32 ok up to here
    t=(-vv+8s+8h)/32 this is incorrect

    \textcolor{red}{\sqrt{v^2-64s+64h} \ne v+8s+8h}
    ...
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  3. #3
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    Quote Originally Posted by mcruz65 View Post
    Code:
    h=16t^2+vt+s (Solve for t)
    h-h=16t^2+vt+s-h
    0=16t^2+vt+s-h
    a=16,b=v,c=s-h
    t= (-b[sqrt](b^2-4ac))/2a
    t=(-v[sqrt]((v)^2-4(16)(s-h) ))/2(16) 
    t=(-v[sqrt](v^2-4(16s-16h) ))/32
    t=(-v[sqrt](v^2-64s+64h))/32
    t=(-vv+8s+8h)/32
    t=(-v)/32 v/32+8s/32+8h/32
    Is this correct??
    Hi mcruz,

    you cannot take the \sqrt{v^2-64s+64h} by taking the square root of each term (almost).

    t=\frac{-v\pm\sqrt{v^2-64(s-h)}}{32}
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