anonymous
  • anonymous
Show that the product of the roots of y=ax^2+bx+c is c/a
Mathematics
schrodinger
  • schrodinger
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johnweldon1993
  • johnweldon1993
Well the roots can be found via the quadratic formula \[\large \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2a}\] The roots would be \(\large \frac{-b + \sqrt{b^2 - 4ac}}{2a}\) and \(\large \frac{-b - \sqrt{b^2 - 4ac}}{2a}\) So lets multiply those 2 \[\large \frac{-b + \sqrt{b^2 - 4ac}}{2a} \times \frac{-b + \sqrt{b^2 - 4ac}}{2a}\] \[\large \frac{(-b + \sqrt{b^2 - 4ac})(-b - \sqrt{b^2 - 4ac})}{4a^2}\] What do you get when you multiply those out?
anonymous
  • anonymous
-b^2/4a^2
johnweldon1993
  • johnweldon1993
Not quite So the top can be distributed out right? Focusing just on the top there \[\large (-b + \sqrt{b^2 - 4ac})(-b - \sqrt{b^2 - 4ac})\] So we have -b times -b = b^2 -b times -√b^2 - 4ac = b√b^2 - 4ac -b times √b^2 - 4ac = -b√b^2 - 4ac and -√b^2 - 4ac times √b^2 - 4ac = -(b^2 - 4ac) So altogether we have *on the top \[\large \frac{b^2 \cancel{-\sqrt{b^2 - 4ac} + \sqrt{b^2 - 4ac}} -b^2 + 4ac}{4a^2}\] or \[\large \frac{b^2 - b^2 + 4ac}{4a^2}\]

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anonymous
  • anonymous
so is the answer going to be 4ac/4a^2
johnweldon1993
  • johnweldon1993
Not quite...if we have \[\large \frac{4ac}{4a^2}\] What can be canceled? hint* we have a 4 on top and bottom...and we have at least 1 'a' on both top and bottom...
anonymous
  • anonymous
ac/a^2
johnweldon1993
  • johnweldon1993
Good...now just the 'a' needs to be taken care of \[\large \frac{\cancel{a}c}{a\cancel{^2}}\] Leaving us with...?
anonymous
  • anonymous
c/a
johnweldon1993
  • johnweldon1993
And that is what we wanted to prove!
anonymous
  • anonymous
can you help me with one more

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