x^2-2x-4 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -2$$
$$c = -4$$
, then
D = b^2 - 4 * a * c =
(-2)^2 - 4 * (1) * (-4) = 20
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 1 + \sqrt{5}$$
$$x_{2} = 1 - \sqrt{5}$$
Vieta's Theorem
it is reduced quadratic equation
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = -4$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 2$$
$$x_{1} x_{2} = -4$$
$$x_{1} = 1 - \sqrt{5}$$
$$x_{2} = 1 + \sqrt{5}$$
Sum and product of roots
[src]
___ ___
1 - \/ 5 + 1 + \/ 5
$$\left(1 - \sqrt{5}\right) + \left(1 + \sqrt{5}\right)$$
$$2$$
/ ___\ / ___\
\1 - \/ 5 /*\1 + \/ 5 /
$$\left(1 - \sqrt{5}\right) \left(1 + \sqrt{5}\right)$$
$$-4$$