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- (7x²-2x+3)-(6x+4x²-1)=-3x²+8x+4
(7x²-2x+3)-(6x+4x²-1)=-3x²+8x-4 equation
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The solution
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[src]
2 2 2 7*x - 2*x + 3 + -6*x - 4*x + 1 = - 3*x + 8*x - 4
$$\left(\left(- 4 x^{2} - 6 x\right) + 1\right) + \left(\left(7 x^{2} - 2 x\right) + 3\right) = \left(- 3 x^{2} + 8 x\right) - 4$$
Detail solution
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$\left(\left(- 4 x^{2} - 6 x\right) + 1\right) + \left(\left(7 x^{2} - 2 x\right) + 3\right) = \left(- 3 x^{2} + 8 x\right) - 4$$
to
$$\left(\left(3 x^{2} - 8 x\right) + 4\right) + \left(\left(\left(- 4 x^{2} - 6 x\right) + 1\right) + \left(\left(7 x^{2} - 2 x\right) + 3\right)\right) = 0$$
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
$$a = 6$$
$$b = -16$$
$$c = 8$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 2$$
$$x_{2} = \frac{2}{3}$$
left part with negative sign.
The equation is transformed from
$$\left(\left(- 4 x^{2} - 6 x\right) + 1\right) + \left(\left(7 x^{2} - 2 x\right) + 3\right) = \left(- 3 x^{2} + 8 x\right) - 4$$
to
$$\left(\left(3 x^{2} - 8 x\right) + 4\right) + \left(\left(\left(- 4 x^{2} - 6 x\right) + 1\right) + \left(\left(7 x^{2} - 2 x\right) + 3\right)\right) = 0$$
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
$$a = 6$$
$$b = -16$$
$$c = 8$$
, then
D = b^2 - 4 * a * c =
(-16)^2 - 4 * (6) * (8) = 64
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 2$$
$$x_{2} = \frac{2}{3}$$
Vieta's Theorem
rewrite the equation
$$\left(\left(- 4 x^{2} - 6 x\right) + 1\right) + \left(\left(7 x^{2} - 2 x\right) + 3\right) = \left(- 3 x^{2} + 8 x\right) - 4$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - \frac{8 x}{3} + \frac{4}{3} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{8}{3}$$
$$q = \frac{c}{a}$$
$$q = \frac{4}{3}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = \frac{8}{3}$$
$$x_{1} x_{2} = \frac{4}{3}$$
$$\left(\left(- 4 x^{2} - 6 x\right) + 1\right) + \left(\left(7 x^{2} - 2 x\right) + 3\right) = \left(- 3 x^{2} + 8 x\right) - 4$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - \frac{8 x}{3} + \frac{4}{3} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{8}{3}$$
$$q = \frac{c}{a}$$
$$q = \frac{4}{3}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = \frac{8}{3}$$
$$x_{1} x_{2} = \frac{4}{3}$$
Sum and product of roots
[src]
sum
2 + 2/3
$$\frac{2}{3} + 2$$
=
8/3
$$\frac{8}{3}$$
product
2*2 --- 3
$$\frac{2 \cdot 2}{3}$$
=
4/3
$$\frac{4}{3}$$
4/3