Question

Derive a formula for the product of the solutions of the quadratic eqaution ax^2 + bx + c in terms of the coefficients

Answer 1

Surely your teacher showed you this -- is it not in your notes or in your textbook?

Answer 2

No it is not

Answer 3

If not, then this geeky website has it for you:
http://www.mathsisfun.com/algebra/quadratic-equation-derivation.html

Answer 4

I know how to derive the equation but I don't understand how to do it for the product of its solutions

Answer 5

Oh sorry, I misread your question.

Answer 6

completeing the square is the long hand version of the quad formula

Answer 7

haha that's okay

Answer 8

or the sum of its solutions for that matter

Answer 9

Now take the two solutions and multiply them together
\[\frac{-b + \sqrt{b^2-4ac}}{2a} \ . \ \frac{-b - \sqrt{b^2-4ac}}{2a} = \frac{b^2 - (b^2 - 4ac)}{4a^2} = \ ....\]

Answer 10

OHHHH!!!!! That makes so much more sense thank you.

Answer 11

\[ax^2+bx+c=0\]
\[x^2+\frac{b}{a}x=-\frac{c}{a}\]
\[x^2+\frac{b}{a}x+(\frac{b}{2a})^2=-\frac{c}{a}+(\frac{b}{2a})^2\]
\[\left(x+\frac{b}{2a}\right)^2=-\frac{c}{a}+\frac{b^2}{4a^2}\]
\[\left(x+\frac{b}{2a}\right)^2=-\frac{4ac+b^2}{4a^2}\]
\[x+\frac{b}{2a}=\pm\sqrt{-\frac{4ac+b^2}{4a^2}}\]
\[x+\frac{b}{2a}=\pm\frac{\sqrt{-4ac+b^2}}{2a}\]
\[x=-\frac{b}{2a}\pm\frac{\sqrt{-4ac+b^2}}{2a}\]
\[x=-\frac{b}{2a}\pm\frac{\sqrt{4ac+b^2}}{2a}\]
\[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]

Answer 12

no I know how to do that, I need a formula for the sum of its two solutions

Answer 13

.... i dropped my negative in the next to last step

Answer 14

So use the quadratic formula and add the two solutions together. You'll see it simplifies nicely.

Answer 15

I got -2b over 2a,
is that correct?

Answer 16

or just -b/a, yes.

Answer 17

yes, thanks so much, it makes a lot of sense now,
what would be correct for the product of the solutions?

Answer 18

Tell me what you think it is and and I'll tell you if it's right.

Answer 19

-4C/4A?

Answer 20

Not quite ...

Answer 21

should "b" be in there?

Answer 22

no

Answer 23

-4ac/4a^2?

Answer 24

NO!, not negative on top

Answer 25

c over a!

Answer 26

Yes

Answer 27

YES!

Answer 28

Now I'm going to show you another way.

Answer 29

wow thanks man your awesome

Answer 30

Suppose ax^2 + bx + c = 0 is a quadratic equation (and hence a is not zero)

Then x^2 + (b/a) x + (c/a) = 0

Answer 31

Let p and q be the roots of this equation.

Then (x - p)(x - q) = 0

Answer 32

i.e., x^2 - (p+q) x + pq = 0

Answer 33

yes, I follow

Answer 34

Comparing the two forms of the equation now, we see that

p + q = -b/a and pq = c/a

Answer 35

The same results we just derived.

Answer 36

oh my gosh, the exact two we just found! That is so cool!

Answer 37

It is cool. Maths works.

Answer 38

haha right, thanks again, I have to go now, but you helped me soooo much!

Answer 39

sure thing. good luck.