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- Equation:
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Equation 3x-5+7x^2=3x^2+7+11x
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Identical expressions
- 3x- five + seven x^ two =3x^ two +7+11x
- 3x minus 5 plus 7x squared equally 3x squared plus 7 plus 11x
- 3x minus five plus seven x to the power of two equally 3x to the power of two plus 7 plus 11x
- 3x-5+7x2=3x2+7+11x
- 3x-5+7x²=3x²+7+11x
- 3x-5+7x to the power of 2=3x to the power of 2+7+11x
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Similar expressions
- 3x-5+7x^2=3x^2+7-11x
- 3x-5+7x^2=3x^2-7+11x
- 3x+5+7x^2=3x^2+7+11x
- 3x-5-7x^2=3x^2+7+11x
3x-5+7x^2=3x^2+7+11x equation
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The solution
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2 2 3*x - 5 + 7*x = 3*x + 7 + 11*x
$$7 x^{2} + \left(3 x - 5\right) = 11 x + \left(3 x^{2} + 7\right)$$
Detail solution
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$7 x^{2} + \left(3 x - 5\right) = 11 x + \left(3 x^{2} + 7\right)$$
to
$$\left(- 11 x + \left(- 3 x^{2} - 7\right)\right) + \left(7 x^{2} + \left(3 x - 5\right)\right) = 0$$
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
$$a = 4$$
$$b = -8$$
$$c = -12$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 3$$
$$x_{2} = -1$$
left part with negative sign.
The equation is transformed from
$$7 x^{2} + \left(3 x - 5\right) = 11 x + \left(3 x^{2} + 7\right)$$
to
$$\left(- 11 x + \left(- 3 x^{2} - 7\right)\right) + \left(7 x^{2} + \left(3 x - 5\right)\right) = 0$$
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
$$a = 4$$
$$b = -8$$
$$c = -12$$
, then
D = b^2 - 4 * a * c =
(-8)^2 - 4 * (4) * (-12) = 256
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 3$$
$$x_{2} = -1$$
Vieta's Theorem
rewrite the equation
$$7 x^{2} + \left(3 x - 5\right) = 11 x + \left(3 x^{2} + 7\right)$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 2 x - 3 = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = -3$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 2$$
$$x_{1} x_{2} = -3$$
$$7 x^{2} + \left(3 x - 5\right) = 11 x + \left(3 x^{2} + 7\right)$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 2 x - 3 = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2$$
$$q = \frac{c}{a}$$
$$q = -3$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 2$$
$$x_{1} x_{2} = -3$$
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