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8 November, 07:58

X^2+10x-7 / x^2+10x+25=0

how do i solve for x?

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Answers (2)
  1. 8 November, 08:15
    0
    You just have to solve the next equation:

    x²+10x-7=0

    Because, if the numerator is equal to "0", then the next expression

    (x²+10x-7) / (x²+10x+25) = 0

    Therefore:

    x²+10x-7=0

    x=[-10⁺₋√ (100+28) ]/2 = (-10⁺₋√128) / 2 = (-10⁺₋8√2) / 2=-5⁺₋4√2

    We have two possible solutions:

    x₁=-5-4√2

    x₂=-5+4√2

    Now, we have to check these possible solutions.

    Remember the denominator cannot be equal to "0".

    if x=-5-4√2

    [ (-5-4√2) ²+10 (-5-4√2) - 7] / [ (-5-4√2) ²+10 (-5-4√2) + 25]=0/32=0

    Therefore: x=-5-4√2 is a solution.

    if x=-5+4√2

    [ (-5+4√2) ²+10 (-5+4√2) - 7] / [ (-5+4√2) ²+10 (-5+4√2) + 25]=0/32=0

    Therefore: x=-5+4√√2 is a solution.

    Answer: we have two solutions for x:

    x₁=-5-4√2

    x₂=-5+4√2
  2. 8 November, 08:19
    0
    Factorize both the numerator and denominator then equate to zero ... you'll get X
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