Question

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

Answer 1

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?

Answer 2

-b^2/4a^2

Answer 3

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

\[\large \frac{b^2 - b^2 + 4ac}{4a^2}\]

Answer 4

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

Answer 5

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...

Answer 6

ac/a^2

Answer 7

Good...now just the 'a' needs to be taken care of

\[\large \frac{\cancel{a}c}{a\cancel{^2}}\]

Leaving us with...?

Answer 8

c/a

Answer 9

And that is what we wanted to prove!

Answer 10

can you help me with one more