Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation (x-5)/2=2*(3x/7-1/2)/3
-
Equation x^4-4*x^3-2*x^2+12*x+9=0
-
Equation sin(x)=-1
-
Equation cos(x)=sqrt(3)/2
- Express {x} in terms of y where:
- 18*x-20*y=-7
- 1*x-6*y=-19
- -19*x-14*y=-8
- 9*x-4*y=2
- Graphing y =:
-
x^4-2*x^2
- Derivative of:
-
x^4-2*x^2
- Limit of the function:
-
x^4-2*x^2
-
Identical expressions
- x^ four - two *x^ two = zero
- x to the power of 4 minus 2 multiply by x squared equally 0
- x to the power of four minus two multiply by x to the power of two equally zero
- x4-2*x2=0
- x⁴-2*x²=0
- x to the power of 4-2*x to the power of 2=0
- x^4-2x^2=0
- x4-2x2=0
- x^4-2*x^2=O
-
Similar expressions
- x^4+2*x^2=0
x^4-2*x^2=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$x^{4} - 2 x^{2} = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 2 v = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -2$$
$$c = 0$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = 2$$
$$v_{2} = 0$$
The final answer:
Because
$$v = x^{2}$$
then
$$x_{1} = \sqrt{v_{1}}$$
$$x_{2} = - \sqrt{v_{1}}$$
$$x_{3} = \sqrt{v_{2}}$$
$$x_{4} = - \sqrt{v_{2}}$$
then:
$$x_{1} = $$
$$\frac{0}{1} + \frac{2^{\frac{1}{2}}}{1} = \sqrt{2}$$
$$x_{2} = $$
$$\frac{\left(-1\right) 2^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{2}$$
$$x_{3} = $$
$$\frac{0^{\frac{1}{2}}}{1} + \frac{0}{1} = 0$$
$$x^{4} - 2 x^{2} = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 2 v = 0$$
This equation is of the form
a*v^2 + b*v + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -2$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(-2)^2 - 4 * (1) * (0) = 4
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 2$$
$$v_{2} = 0$$
The final answer:
Because
$$v = x^{2}$$
then
$$x_{1} = \sqrt{v_{1}}$$
$$x_{2} = - \sqrt{v_{1}}$$
$$x_{3} = \sqrt{v_{2}}$$
$$x_{4} = - \sqrt{v_{2}}$$
then:
$$x_{1} = $$
$$\frac{0}{1} + \frac{2^{\frac{1}{2}}}{1} = \sqrt{2}$$
$$x_{2} = $$
$$\frac{\left(-1\right) 2^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{2}$$
$$x_{3} = $$
$$\frac{0^{\frac{1}{2}}}{1} + \frac{0}{1} = 0$$
Rapid solution
[src]
x1 = 0
$$x_{1} = 0$$
___ x2 = -\/ 2
$$x_{2} = - \sqrt{2}$$
___ x3 = \/ 2
$$x_{3} = \sqrt{2}$$
x3 = sqrt(2)
Sum and product of roots
[src]
sum
___ ___ - \/ 2 + \/ 2
$$- \sqrt{2} + \sqrt{2}$$
=
0
$$0$$
product
/ ___\ ___ 0*\-\/ 2 /*\/ 2
$$\sqrt{2} \cdot 0 \left(- \sqrt{2}\right)$$
=
0
$$0$$
0
The graph