Question regarding quadratic equation roots

Hello guys,

I was wondering that since the product of the two roots of a quadratic equation equal to the y-intercept (c) in the quadratic form and that the sum of the two roots equal to (-b) in the quadratic form.

Does anyone know why or how this was invented?

Hello guys,

I was wondering that since the product of the two roots of a quadratic equation equal to the y-intercept (c) in the quadratic form and that the sum of the two roots equal to (-b) in the quadratic form.

Does anyone know why or how this was invented?

Also a correction:

The product of the two roots is equal to $\frac{c}{a}$

and the sum is equal to $-\frac{b}{a}$.

A quadratic can be written in the form

$a(x-p)(x-q)$

where $p$ and $q$ are the roots of the quadratic.

Expand this and compare it with $ax^2+bx+c$.

Is that the actual way to prove the theorem - that the products are equal to +c

It's one way. The bear has given you another way.

you could also try adding and multiplying the two versions ( one with +, one with - ) of the quadratic formula

A quadratic can be written in the form

$a(x-p)(x-q)$

where $p$ and $q$ are the roots of the quadratic.

Expand this and compare it with $ax^2+bx+c$.

a(x-p)(x-q)
a(x^2-xq-px+pq)
ax^2 - axq - apx + apq = 0
ax^2 + bx + c = 0

apq = c
pq = c/a

- axq - apx = bx
- xq - px = bx/a
- q - p = b/a
q + p = -b/a

Is this correct?

Done in head and on iPad so sorry if there are any spelling mistakes lol.
Posted from TSR Mobile

Edit: Actually not quite. You can compare the coefficients, e.g. -p - q = b/a etc but not the complete terms (i.e. no -px - qx = -bx/a).

As an exercise you might now try to find similar results for a cubic equation.

Sorry I don't get what you mean by the first line :/, especially the complete terms part.

Posted from TSR Mobile

You are correct that $ax^2 + bx + c \equiv ax^2 - apx - aqx + apq$ {Note that the use of $\equiv$ means that it holds whatever value of x you substitute in. It means "is the same as" or "is equivalent to"}

From here we can do what is called "comparing coefficients" which is exactly what is says on the tin. We can compare the coefficients of x to see that $- ap - aq = b$ and hence $p + q = \frac{-b}{a}$. We would not write $-apx - aqx = bx$, we would only write the coefficients down.

x

Yes.

Just to make it easier for you to see, if we swapped x-1 for a then we would have a^3 = a(1+x).

So then a^3 - a(1+x) = 0 and there is a common factor between those two terms.

That saves you a bit of expanding.

So, x^3 = x(1+x).
x^3 - x(1+x) = 0
x^3 - x^2 - x = 0
x(x^2 - x - 1) = 0

Then what?

Oh, you should keep x-1 in there instead of putting a (which you've now changed into x which is very dangerous!). I just labelled it a to make it clear that you were factoring out.

a^3 - a(1+x) = 0
You've then multiplied it out and then factored it out which is fine but you can skip the third line by just noticing that a is the commmon factor, so a^3 - a(1+x) = a(a^2 - (1+x)), and then substitute a back out and factorise whatever's in the brackets. Ideally you don't need to use 'a', just use (x-1) from the start and notice that you can factor out (x-1) from (x-1)^3 - (x-1)(x+1).

$So, (x-1)^3 - (x-1)(1+x) = 0[br](x-1)^2(1+x) = 0[br](x^2-2x+1)(1+x) = 0[br](x^3 - 2x^2 + x) + (x^2 - 2x + 1) = 0[br]x^3 - x^2 - x + 1 = 0$

This is clearly incorrect.

V2

$(x-1)^3 - (x-1)(1+x) = 0[br](x-1)^2 - (1+x) = 0[br](x^2 - 2x + 1) - 1 - x = 0[br]x^2 - 3x = 0[br]x(x-3) = 0[br]x = 0 or x = 3$

Partially correct, where is the x = 1 gone?