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3*x^2-18*x+27=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
This equation is of the form
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
$$a = 3$$
$$b = -18$$
$$c = 27$$
, then
Because D = 0, then the equation has one root.
$$x_{1} = 3$$
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
$$a = 3$$
$$b = -18$$
$$c = 27$$
, then
D = b^2 - 4 * a * c =
(-18)^2 - 4 * (3) * (27) = 0
Because D = 0, then the equation has one root.
x = -b/2a = --18/2/(3)
$$x_{1} = 3$$
Vieta's Theorem
rewrite the equation
$$\left(3 x^{2} - 18 x\right) + 27 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 6 x + 9 = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -6$$
$$q = \frac{c}{a}$$
$$q = 9$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 6$$
$$x_{1} x_{2} = 9$$
$$\left(3 x^{2} - 18 x\right) + 27 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 6 x + 9 = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -6$$
$$q = \frac{c}{a}$$
$$q = 9$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 6$$
$$x_{1} x_{2} = 9$$
The graph