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How to use it?
- Equation:
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Equation 16-3x^2+14x=5x-14
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Equation cos(pi*x)=0
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Equation tg(x)=3
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Identical expressions
- sixteen -3x^ two + fourteen x=5x-14
- 16 minus 3x squared plus 14x equally 5x minus 14
- sixteen minus 3x to the power of two plus fourteen x equally 5x minus 14
- 16-3x2+14x=5x-14
- 16-3x²+14x=5x-14
- 16-3x to the power of 2+14x=5x-14
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Similar expressions
- 16-3x^2-14x=5x-14
- 16+3x^2+14x=5x-14
- 16-3x^2+14x=5x+14
16-3x^2+14x=5x-14 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$14 x + \left(16 - 3 x^{2}\right) = 5 x - 14$$
to
$$\left(14 - 5 x\right) + \left(14 x + \left(16 - 3 x^{2}\right)\right) = 0$$
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
$$a = -3$$
$$b = 9$$
$$c = 30$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = -2$$
$$x_{2} = 5$$
left part with negative sign.
The equation is transformed from
$$14 x + \left(16 - 3 x^{2}\right) = 5 x - 14$$
to
$$\left(14 - 5 x\right) + \left(14 x + \left(16 - 3 x^{2}\right)\right) = 0$$
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
$$a = -3$$
$$b = 9$$
$$c = 30$$
, then
D = b^2 - 4 * a * c =
(9)^2 - 4 * (-3) * (30) = 441
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = -2$$
$$x_{2} = 5$$
Vieta's Theorem
rewrite the equation
$$14 x + \left(16 - 3 x^{2}\right) = 5 x - 14$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 3 x - 10 = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -3$$
$$q = \frac{c}{a}$$
$$q = -10$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 3$$
$$x_{1} x_{2} = -10$$
$$14 x + \left(16 - 3 x^{2}\right) = 5 x - 14$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 3 x - 10 = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -3$$
$$q = \frac{c}{a}$$
$$q = -10$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 3$$
$$x_{1} x_{2} = -10$$
Sum and product of roots
[src]
sum
-2 + 5
$$-2 + 5$$
=
3
$$3$$
product
-2*5
$$- 10$$
=
-10
$$-10$$
-10
The graph