Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation x^2+x=12
-
Equation 3x-2(3x-2)=19
-
Equation x^4=(x-56)^2
-
Equation (-2x+1)*(-2x-7)=0
- Express {x} in terms of y where:
- -7*x-13*y=-12
- 8*x+1*y=4
- 20*x-16*y=7
- -3*x+19*y=16
- Derivative of:
-
5x^4+2x^3
- Integral of d{x}:
-
5x^4+2x^3
-
Identical expressions
- 5x^ four +2x^ three
- 5x to the power of 4 plus 2x cubed
- 5x to the power of four plus 2x to the power of three
- 5x4+2x3
- 5x⁴+2x³
- 5x to the power of 4+2x to the power of 3
-
Similar expressions
- 5x^4-2x^3
5x^4+2x^3 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$5 x^{4} + 2 x^{3} = 0$$
transform
Take common factor x^2 from the equation
we get:
$$x^{2} \left(5 x^{2} + 2 x\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$5 x^{2} + 2 x = 0$$
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
$$a = 5$$
$$b = 2$$
$$c = 0$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = 0$$
$$x_{3} = - \frac{2}{5}$$
The final answer for 5*x^4 + 2*x^3 = 0:
$$x_{1} = 0$$
$$x_{2} = 0$$
$$x_{3} = - \frac{2}{5}$$
$$5 x^{4} + 2 x^{3} = 0$$
transform
Take common factor x^2 from the equation
we get:
$$x^{2} \left(5 x^{2} + 2 x\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$5 x^{2} + 2 x = 0$$
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
$$a = 5$$
$$b = 2$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(2)^2 - 4 * (5) * (0) = 4
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = 0$$
$$x_{3} = - \frac{2}{5}$$
The final answer for 5*x^4 + 2*x^3 = 0:
$$x_{1} = 0$$
$$x_{2} = 0$$
$$x_{3} = - \frac{2}{5}$$
Sum and product of roots
[src]
sum
-2/5
$$- \frac{2}{5}$$
=
-2/5
$$- \frac{2}{5}$$
product
0*(-2) ------ 5
$$\frac{\left(-2\right) 0}{5}$$
=
0
$$0$$
0