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

1. 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?

2. anonymous

-b^2/4a^2

3. 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 $\large \frac{b^2 - b^2 + 4ac}{4a^2}$

4. anonymous

so is the answer going to be 4ac/4a^2

5. 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...

6. anonymous

ac/a^2

7. johnweldon1993

Good...now just the 'a' needs to be taken care of $\large \frac{\cancel{a}c}{a\cancel{^2}}$ Leaving us with...?

8. anonymous

c/a

9. johnweldon1993

And that is what we wanted to prove!

10. anonymous

can you help me with one more