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stupidinmath Group Title

Find all the integers m for which y^2+my+50 can be factored.

  • one year ago
  • one year ago

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  1. EulerGroupie Group Title
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    When factoring this form of expression, you are looking for two numbers whose product is 50 and sum is m. 50 is positive so it will take multiplying either two positive numbers or two negative numbers to make it positive. Start by finding all of the pairs of factors that make 50. pos factors neg factors pos sum neg sum 1 50 -1 -50 51 -51 2 25 -2 -25 27 -27 5 10 -5 -10 15 -15 m can be anythin in the pos or neg sum columns above.

    • one year ago
  2. mukushla Group Title
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    allow me to do some thinkin on this question : Discriminant of quadratic must be a complete square\[m^2-100=n^2\]\[(m-n)(m+n)=100\]

    • one year ago
  3. EulerGroupie Group Title
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    If the discriminant is b^2-4ac isn't it m^2-4(50)? \[m ^{2}-200=n ^{2}\]\[(m-n)(m+n)=200\]I would like to see where you are going with this. :)

    • one year ago
  4. mukushla Group Title
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    oh sorry thats it...

    • one year ago
  5. mukushla Group Title
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    now going to solve it for positive m,n's like this\[200=2\times100\]\[200=4\times50\]... note that \(m-n\) and \(m+n\) both are even for the first one for example \(m-n=2\) , \(m+n=100\) it gives \(m=51\)

    • one year ago
  6. mukushla Group Title
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    another one gives m=27

    • one year ago
  7. punnus Group Title
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    This is a Quadratic equation. The general form of a given quadratic equation is \[ax ^{2}+bx+c=0\] Now for solution to this equation we do this \[b ^{2}-4ac=0\] for equal solutions So we get two values that are \[+10\sqrt{2}\]and

    • one year ago
  8. punnus Group Title
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    \[-10\sqrt{2}\]

    • one year ago
  9. punnus Group Title
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    these both are values for m when the solution of this function will form a cusp on the y axis

    • one year ago
  10. mukushla Group Title
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    \[200=10\times 20\]...

    • one year ago
  11. mukushla Group Title
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    and we can find all possibilities...also for every positive m its negative is an answer for us

    • one year ago
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