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anonymous
 one year ago
Show that the product of the roots of y=ax^2+bx+c is c/a
anonymous
 one year ago
Show that the product of the roots of y=ax^2+bx+c is c/a

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johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.0Well 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?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.0Not 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 \[\large \frac{b^2  b^2 + 4ac}{4a^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so is the answer going to be 4ac/4a^2

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.0Not 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...

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.0Good...now just the 'a' needs to be taken care of \[\large \frac{\cancel{a}c}{a\cancel{^2}}\] Leaving us with...?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.0And that is what we wanted to prove!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you help me with one more
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