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anonymous

  • 5 years ago

4x^2 +24x+35=0

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  1. anonymous
    • 5 years ago
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    you need to factor...

  2. anonymous
    • 5 years ago
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    x = \[-24 \pm \sqrt{24^2-4(4)(35) }\] \[/2(4)\]

  3. anonymous
    • 5 years ago
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    good call, use the quadratic equation for that. You could, however, factor that. 35 has the following primes: 5, 7. 24 has the following primes: 2,2,2,3, and 4 has two primes, 2,2. Don't think this will factor into easy integers, so just use quadratic equation here.

  4. anonymous
    • 5 years ago
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    Doesn't look like it factors easily, so use completing the square: \[4x ^2+24x +35 = 0\] (Divide by the x-squared coeficient) \[x ^2+6x +35/4 = 0\] (Move the constant term to the other side) \[x ^2+6x = -35/4\] (Take half the coeficient of the x term, square it, and add to both sides) \[x ^2+6x +3^2 = -35/4 +3^2\] (Now the left side of the equation is a perfect square, and factors easily) \[(x +3)^2 = -35/4 +9\] (take the square root of each side) \[x +3 = \pm\sqrt{-35/4 +9}\] (now just move the three over, and you're done) \[x =-3 \pm\sqrt{-35/4 +9}\]

  5. anonymous
    • 5 years ago
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    Ends up reducing to -5/2 and -7/2

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