Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation (x+2)/9=(x-3)/2
-
Equation -x-4+5(x+3)=5(-1-x)-2
-
Equation 3(x+3)=2x+5
-
Equation x^2-5x+4=0
- Express {x} in terms of y where:
- 17*x-20*y=6
- -11*x+15*y=7
- 14*x-3*y=-5
- 2*x+3*y=14
- Graphing y =:
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x^4-x^2
- Derivative of:
-
x^4-x^2
- Factor polynomial:
- x^4-x^2
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Identical expressions
- x^ four -x^ two = zero
- x to the power of 4 minus x squared equally 0
- x to the power of four minus x to the power of two equally zero
- x4-x2=0
- x⁴-x²=0
- x to the power of 4-x to the power of 2=0
- x^4-x^2=O
-
Similar expressions
- x^4+x^2=0
x^4-x^2=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$x^{4} - x^{2} = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - v = 0$$
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
$$a = 1$$
$$b = -1$$
$$c = 0$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = 1$$
$$v_{2} = 0$$
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:
$$x_{1} = $$
$$\frac{0}{1} + \frac{1^{\frac{1}{2}}}{1} = 1$$
$$x_{2} = $$
$$\frac{\left(-1\right) 1^{\frac{1}{2}}}{1} + \frac{0}{1} = -1$$
$$x_{3} = $$
$$\frac{0^{\frac{1}{2}}}{1} + \frac{0}{1} = 0$$
$$x^{4} - x^{2} = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - v = 0$$
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
$$a = 1$$
$$b = -1$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(-1)^2 - 4 * (1) * (0) = 1
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 1$$
$$v_{2} = 0$$
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:
$$x_{1} = $$
$$\frac{0}{1} + \frac{1^{\frac{1}{2}}}{1} = 1$$
$$x_{2} = $$
$$\frac{\left(-1\right) 1^{\frac{1}{2}}}{1} + \frac{0}{1} = -1$$
$$x_{3} = $$
$$\frac{0^{\frac{1}{2}}}{1} + \frac{0}{1} = 0$$
The graph