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Identical expressions
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Similar expressions
- 49x^3-14x^2+x=0
- 49x^3+14x^2-x=0
49x^3+14x^2+x=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$49 x^{3} + 14 x^{2} + x = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(49 x^{2} + 14 x + 1\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$49 x^{2} + 14 x + 1 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 49$$
$$b = 14$$
$$c = 1$$
, then
Because D = 0, then the equation has one root.
$$x_{2} = - \frac{1}{7}$$
The final answer for (49*x^3 + 14*x^2 + x) + 0 = 0:
$$x_{1} = 0$$
$$x_{2} = - \frac{1}{7}$$
$$49 x^{3} + 14 x^{2} + x = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(49 x^{2} + 14 x + 1\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$49 x^{2} + 14 x + 1 = 0$$
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:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 49$$
$$b = 14$$
$$c = 1$$
, then
D = b^2 - 4 * a * c =
(14)^2 - 4 * (49) * (1) = 0
Because D = 0, then the equation has one root.
x = -b/2a = -14/2/(49)
$$x_{2} = - \frac{1}{7}$$
The final answer for (49*x^3 + 14*x^2 + x) + 0 = 0:
$$x_{1} = 0$$
$$x_{2} = - \frac{1}{7}$$
Vieta's Theorem
rewrite the equation
$$49 x^{3} + 14 x^{2} + x = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} + \frac{2 x^{2}}{7} + \frac{x}{49} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{2}{7}$$
$$q = \frac{c}{a}$$
$$q = \frac{1}{49}$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = - \frac{2}{7}$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = \frac{1}{49}$$
$$x_{1} x_{2} x_{3} = 0$$
$$49 x^{3} + 14 x^{2} + x = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} + \frac{2 x^{2}}{7} + \frac{x}{49} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{2}{7}$$
$$q = \frac{c}{a}$$
$$q = \frac{1}{49}$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = - \frac{2}{7}$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = \frac{1}{49}$$
$$x_{1} x_{2} x_{3} = 0$$
Sum and product of roots
[src]
sum
0 - 1/7 + 0
$$\left(- \frac{1}{7} + 0\right) + 0$$
=
-1/7
$$- \frac{1}{7}$$
product
1*-1/7*0
$$1 \left(- \frac{1}{7}\right) 0$$
=
0
$$0$$
0
The graph