Basic Venn diagram probablility.

I'm taking A level mathematics and futher mathematics, but I'm struggling a little when it comes to probability. I ended up solving the question in the images, but I am unsure why my convoluted first method didn't work. So my question is why didn't my original mehtod work? The question and my original workiare provided in the image.

Probability
Mathgnome Mathgnome
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Matchmaticians Basic Venn diagram probablility. File #1 File #1 (png)
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  • Mathe Mathe
    0

    Would you consider increasing the bounty? $5 is very little for our time.

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Mathe Mathe
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  • Mathgnome Mathgnome
    0

    Firstly, how did I complete the square wrong? (c + 1/5)^2 + (16/5)c -> c^2 + (2/5)c + 1/25 + (16/5)c -> c^2 + (18/5)c + 1/25 -> (c + 9/5)^2 + 1/25 - 81/25 -> c^2 + (18/5)c - 16/5. Lastly , I was assuming that the question only had one solution (which I now realise was somethign that I should never do) so I thought that the discriminent in that case would be zero which would allow you to solve for the two unknowns.

    • Mathe Mathe
      +1

      1/25 - 81/25 = -80/25 but you wrote down something else.

    • Mathgnome Mathgnome
      0

      I didn't mean to accept the answer. Could you repley to a few of my message? I will tip 5$.

    • Mathe Mathe
      +1

      Sure, I will reply. Go ahead.

    • Mathe Mathe
      +1

      I see now that -80/25 = - 16/5, so that was actually correct!

  • Mathgnome Mathgnome
    0

    Thanks for the clarification I thought I was going crazy. So when you said that you can't solve for two unkowns with a single equation I do know that, but I thought the question was saying that there was only one solution for each (which there is), so surely creating a quadratic equation and setting the discriminent equal to zero would mean that it only has one root and therefore only one answer. So, my question is why didn't it work.

    • Mathe Mathe
      +1

      You could have one root to the quadratic equation where both c and d are positive (true solution) and another root to the quadratic equation were both c and d are negative (not real solutions). This is probably the case. You could replace the values ofc and d that you found on the quadratic equation and see that they must achieve a zero.

  • Mathgnome Mathgnome
    0

    Okay thanks so much that really clears it up for me. I not sure why, but for some reason I was thinking that because there was only one solution to the question I was thinking that if I manipulated it as shown in the images to a quadratic equation the quadratic equation would only have one solution because the question only had one solution.

    • Mathe Mathe
      +1

      There are very, very few cases where you can solve two unknowns with only one equation.

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