why is the sum of the roots of a quadratic equation -b/a?and the product c/a?how do we know that it always is?

Question

i need some serious help- no site has it and its upsetting.

Answer 1

The quadratic formula is:

x = [ -b +/- sqrt(b² - 4ac) ] / 2a

So the two roots are:
[-b + sqrt(b² - 4ac) ] / 2a
[-b - sqrt(b² - 4ac) ] / 2a

To add these together, just add the numerators, since they have the same denominator.
[ -b + sqrt(b² - 4ac) + -b - sqrt(b² - 4ac) ] / 2a

Notice how the sqrt cancels out.
(-b + -b ) / 2a
= -2b / 2a

Then a 2 cancels out:
= -b / a

You can do a similar thing for the product.

For the two roots, multiply them together:
[-b + sqrt(b² - 4ac) ] / 2a * [-b - sqrt(b² - 4ac) ] / 2a

Remember to multiply two fractions, just multiply the numerator across, and also multiply the denominator across.
( [-b + sqrt(b² - 4ac) ] * [-b - sqrt(b² - 4ac) ] / ( 2a * 2a )

Whenever you have (X + Y)(X - Y), remember that outer and inner terms (XY and -XY) cancel out. Thus it is just X² - Y²

So this simplifies to:
[ (-b)² - (sqrt(b² - 4ac))² ] / 4a²

This further simplifies to:
[ b² - (b² - 4ac) ] / 4a²

Then the b² cancels out:
4ac / 4a²

Now cancel 4a from top and bottom:
c / a

So the sum of the roots is -b / a
And the product of the roots is c / a

Answer 2

The answer lies in the general solution to a quadratic equation, which is usually just termed "the quadratic equation." This gives the general solution to the quadratic equation:

a * x^2 + b * x + c = 0

as: x1 = (-b + sqrt(b^2 - 4*a*c))/2a and x2 = (-b - sqrt(b^2 - 4*a*c))/2a

Do the algebra and you will see that x1 + x2 = -b/a and x1*x2 = c/a

Answer 3

Well, the quadratic formula (general solution for a quadratic equation) is

[-b +- sqrt(b^2 - 4ac)]/2a
giving two roots - adding them together = -2b/2a = -b/a

Multiplying them together = (b^2 - b^2 + 4ac)/4a^2 = c/a

Answer 4

This falls out of the usual formula for solving a quadratic equation. It is fairly easy to derive (and thus prove) this formula.

Answer 5

This comes from the quadratic formula:

x1=(-b+√(b²-4ac))/(2a), x2=(-b-√(b²-4ac))/(2a)
x1+x2 = (-b - b + √(b²-4ac) - √(b²-4ac))/(2a) = -2b/(2a) = -b/a
x1*x2 = (b² + b√(b²-4ac) - b√(b²-4ac) - (b²-4ac))/(4a²) = (b²-b² + 4ac)/(4a²) = 4ac/(4a²) = c/a

The quadratic formula itself may be proven from the process of completing the sqauare:

ax² + bx + c = 0
x² + bx/a + c/a = 0
x² + bx/a = -c/a
x² + bx/a + (b/(2a))² = -c/a + (b/(2a))²
(x + b/(2a))² = -c/a + (b/(2a))²
(x + b/(2a))² = -c/a + b²/(4a²)
(x + b/(2a))² = -4ac/(4a²) + b²/(4a²)
(x + b/(2a))² = (b²-4ac)/(4a²)
x + b/(2a) = ±√(b²-4ac)/(2a)
x=(-b±√(b²-4ac))/(2a)

Answer 6

http://www.phar.cam.ac.uk/docs/emb/sbh_quad.pdf might help.