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Solve the equation and show the check of the potential answer(s). If any answers are excluded values, state this on your answer sheet.

a2 + 4a - 5/a2 + 2a - a + 3/a + 2 =1/a2 + 2a

Can someone please show me how to do this problem with checks? I have submitted it to my homework three times and cannot get the answer correct.

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4 answers

  1. Can't help, I have no idea what the expression is. Is it this?
    (a^2 + 4a - 5)/(a^2 + 2a - a) + 3/(a + 2 )=1/(a^2 + 2a )

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  2. Yes that's the one Bob

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  3. (a^2 + 4a - 5)/(a^2 + 2a - a) + 3/(a + 2 )=1/(a^2 + 2a )
    (a+5)(a-1)/( a^2+2a-a) + 3/(a+2) = 1/a(a+2)

    I Suspect at this point, you do not have the problem correct, the (a^2+2a-a) term looks suspiciously wrong

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  4. Looking at the distribution of denominators of (a-1), (a+2), a, and another (a+2), is suspect that mysterious denominator is (a+2)(a-1) or (a^2 + a - 2)

    so
    (a+5)(a-1)/( a^2 + a -2) + 3/(a+2) = 1/a(a+2)
    (a+5)(a-1)/((a + 2)(a - 1) + 3/(a+2) = 1/(a(a+2) )
    (a+5)/(a + 2) + 3/(a+2) = 1/a(a+2)
    multiply each term by a(a+2)
    a(a+5) + 3a = 1
    a^2 + 5a + 3a - 1 = 0
    a^2 + 8a = 1
    completing the square:
    a^2 + 8a + 16 = 1 + 16
    (a+4)^2 = 17
    a+4 = ± √17
    a = -4 ± √17 <------ based on my assumption

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