Mister Exam
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How to use it?
- Equation:
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Equation x^4-2x^2+1=0
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Equation x^3=125
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- Derivative of:
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x^4-2x^2+1
- Graphing y =:
- x^4-2x^2+1
-
Identical expressions
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- x^4-2x^2+1=O
-
Similar expressions
- x^4+2x^2+1=0
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- x^4-2x^2-1=0
x^4-2x^2+1=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$x^{4} - 2 x^{2} + 1 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 2 v + 1 = 0$$
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
$$a = 1$$
$$b = -2$$
$$c = 1$$
, then
Because D = 0, then the equation has one root.
$$v_{1} = 1$$
The final answer:
Because
$$v = x^{2}$$
then
$$x_{1} = \sqrt{v_{1}}$$
$$x_{2} = - \sqrt{v_{1}}$$
then:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 1^{\frac{1}{2}}}{1} = 1$$
$$x_{2} = \frac{\left(-1\right) 1^{\frac{1}{2}}}{1} + \frac{0}{1} = -1$$
$$x^{4} - 2 x^{2} + 1 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 2 v + 1 = 0$$
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
$$a = 1$$
$$b = -2$$
$$c = 1$$
, then
D = b^2 - 4 * a * c =
(-2)^2 - 4 * (1) * (1) = 0
Because D = 0, then the equation has one root.
v = -b/2a = --2/2/(1)
$$v_{1} = 1$$
The final answer:
Because
$$v = x^{2}$$
then
$$x_{1} = \sqrt{v_{1}}$$
$$x_{2} = - \sqrt{v_{1}}$$
then:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 1^{\frac{1}{2}}}{1} = 1$$
$$x_{2} = \frac{\left(-1\right) 1^{\frac{1}{2}}}{1} + \frac{0}{1} = -1$$
Sum and product of roots
[src]
sum
0 - 1 + 1
$$\left(-1 + 0\right) + 1$$
=
0
$$0$$
product
1*-1*1
$$1 \left(-1\right) 1$$
=
-1
$$-1$$
-1
The graph