Mister Exam
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How to use it?
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Identical expressions
- four x^4+5x^ two + one = zero
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Similar expressions
- 4x^4-5x^2+1=0
- 4x^4+5x^2-1=0
4x^4+5x^2+1=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\left(4 x^{4} + 5 x^{2}\right) + 1 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$4 v^{2} + 5 v + 1 = 0$$
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
$$a = 4$$
$$b = 5$$
$$c = 1$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = - \frac{1}{4}$$
$$v_{2} = -1$$
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:
$$x_{1} = $$
$$\frac{0}{1} + \frac{\left(- \frac{1}{4}\right)^{\frac{1}{2}}}{1} = \frac{i}{2}$$
$$x_{2} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(- \frac{1}{4}\right)^{\frac{1}{2}}}{1} = - \frac{i}{2}$$
$$x_{3} = $$
$$\frac{0}{1} + \frac{\left(-1\right)^{\frac{1}{2}}}{1} = i$$
$$x_{4} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(-1\right)^{\frac{1}{2}}}{1} = - i$$
$$\left(4 x^{4} + 5 x^{2}\right) + 1 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$4 v^{2} + 5 v + 1 = 0$$
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
$$a = 4$$
$$b = 5$$
$$c = 1$$
, then
D = b^2 - 4 * a * c =
(5)^2 - 4 * (4) * (1) = 9
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = - \frac{1}{4}$$
$$v_{2} = -1$$
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:
$$x_{1} = $$
$$\frac{0}{1} + \frac{\left(- \frac{1}{4}\right)^{\frac{1}{2}}}{1} = \frac{i}{2}$$
$$x_{2} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(- \frac{1}{4}\right)^{\frac{1}{2}}}{1} = - \frac{i}{2}$$
$$x_{3} = $$
$$\frac{0}{1} + \frac{\left(-1\right)^{\frac{1}{2}}}{1} = i$$
$$x_{4} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(-1\right)^{\frac{1}{2}}}{1} = - i$$
Sum and product of roots
[src]
sum
I I
-I - - + - + I
2 2
$$\left(\left(- i - \frac{i}{2}\right) + \frac{i}{2}\right) + i$$
=
0
$$0$$
product
-I I
-I*---*-*I
2 2
$$i \frac{i}{2} \cdot - i \left(- \frac{i}{2}\right)$$
=
1/4
$$\frac{1}{4}$$
1/4
Rapid solution
[src]
x1 = -I
$$x_{1} = - i$$
-I
x2 = ---
2
$$x_{2} = - \frac{i}{2}$$
I
x3 = -
2
$$x_{3} = \frac{i}{2}$$
x4 = I
$$x_{4} = i$$
x4 = i