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.

  • 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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  • 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.

    • 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!

  • 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.

  • 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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