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anonymous

  • 5 years ago

need to find 2 solutions to this equation

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  1. anonymous
    • 5 years ago
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    \[(3x+8)^{2}-4(3x+8)-77=0\]

  2. anonymous
    • 5 years ago
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    Let k = (3x+8) So the above equation becomes \[k^2 - 4k - 77 = 0\] Solve the quadratic for two solutions of k, then replace k in those two equations with 3x+8 and solve for your two x solutions.

  3. anonymous
    • 5 years ago
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    ok thanks, I will try that out. Not confident I'll get it right so you can correct me if it's wrong :)

  4. anonymous
    • 5 years ago
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    i'm not sure if i'm doing this step correct. I got -76 for one..?

  5. anonymous
    • 5 years ago
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    \[k = \frac{4 \pm \sqrt{16 + 4*77}}{2}\] \[k = 2 \pm \sqrt{324}\] \[k = 2 \pm 18\] \[k = -16 \text{ or } k = 20\] Therefore \[3x+8 = -16\] or \[3x+8 = 20\]

  6. anonymous
    • 5 years ago
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    So working with the first equation: \[3x + 8 = -16\] \[3x = -24\] \[x = -8\]

  7. anonymous
    • 5 years ago
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    Working the second one: \[3x+8 = 20\] \[3x = 12\] \[x = 4\]

  8. anonymous
    • 5 years ago
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    So our two solutions are \(x \in \{-8,4\}\)

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