Mister Exam
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- (-3x^2+6x-16)=0
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(-3x^2+6x+16)=0 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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = -3$$
$$b = 6$$
$$c = 16$$
, then
$$D = b^2 - 4\ a\ c = $$
$$6^{2} - \left(-3\right) 4 \cdot 16 = 228$$
Because D > 0, then the equation has two roots.
$$x_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{1} = - \frac{\sqrt{57}}{3} + 1$$
Simplify
$$x_{2} = 1 + \frac{\sqrt{57}}{3}$$
Simplify
$$a\ x^2 + b\ x + c = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = -3$$
$$b = 6$$
$$c = 16$$
, then
$$D = b^2 - 4\ a\ c = $$
$$6^{2} - \left(-3\right) 4 \cdot 16 = 228$$
Because D > 0, then the equation has two roots.
$$x_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{1} = - \frac{\sqrt{57}}{3} + 1$$
Simplify
$$x_{2} = 1 + \frac{\sqrt{57}}{3}$$
Simplify
Vieta's Theorem
rewrite the equation
$$- 3 x^{2} + 6 x + 16 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 2 x - \frac{16}{3} = 0$$
$$p x + x^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = - \frac{16}{3}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 2$$
$$x_{1} x_{2} = - \frac{16}{3}$$
$$- 3 x^{2} + 6 x + 16 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 2 x - \frac{16}{3} = 0$$
$$p x + x^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = - \frac{16}{3}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 2$$
$$x_{1} x_{2} = - \frac{16}{3}$$
Sum and product of roots
[src]
sum
____ ____
\/ 57 \/ 57
1 - ------ + 1 + ------
3 3
$$\left(- \frac{\sqrt{57}}{3} + 1\right) + \left(1 + \frac{\sqrt{57}}{3}\right)$$
=
2
$$2$$
product
____ ____
\/ 57 \/ 57
1 - ------ * 1 + ------
3 3
$$\left(- \frac{\sqrt{57}}{3} + 1\right) * \left(1 + \frac{\sqrt{57}}{3}\right)$$
=
-16/3
$$- \frac{16}{3}$$
Rapid solution
[src]
____
\/ 57
x_1 = 1 - ------
3
$$x_{1} = - \frac{\sqrt{57}}{3} + 1$$
____
\/ 57
x_2 = 1 + ------
3
$$x_{2} = 1 + \frac{\sqrt{57}}{3}$$
The graph