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- 16x+5x^2-12=0
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16x+5x^2+12=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
$$a = 5$$
$$b = 16$$
$$c = 12$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 5 \cdot 4 \cdot 12 + 16^{2} = 16$$
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
$$x_{1} = - \frac{6}{5}$$
Simplify
$$x_{2} = -2$$
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
$$a = 5$$
$$b = 16$$
$$c = 12$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 5 \cdot 4 \cdot 12 + 16^{2} = 16$$
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
$$x_{1} = - \frac{6}{5}$$
Simplify
$$x_{2} = -2$$
Simplify
Vieta's Theorem
rewrite the equation
$$5 x^{2} + 16 x + 12 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} + \frac{16 x}{5} + \frac{12}{5} = 0$$
$$p x + x^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{16}{5}$$
$$q = \frac{c}{a}$$
$$q = \frac{12}{5}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = - \frac{16}{5}$$
$$x_{1} x_{2} = \frac{12}{5}$$
$$5 x^{2} + 16 x + 12 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} + \frac{16 x}{5} + \frac{12}{5} = 0$$
$$p x + x^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{16}{5}$$
$$q = \frac{c}{a}$$
$$q = \frac{12}{5}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = - \frac{16}{5}$$
$$x_{1} x_{2} = \frac{12}{5}$$
Sum and product of roots
[src]
sum
-2 + -6/5
$$\left(-2\right) + \left(- \frac{6}{5}\right)$$
=
-16/5
$$- \frac{16}{5}$$
product
-2 * -6/5
$$\left(-2\right) * \left(- \frac{6}{5}\right)$$
=
12/5
$$\frac{12}{5}$$
The graph