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x^4+2x^2-8=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$x^{4} + 2 x^{2} - 8 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} + 2 v - 8 = 0$$
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
$$a = 1$$
$$b = 2$$
$$c = -8$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = 2$$
Simplify
$$v_{2} = -4$$
Simplify
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:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 2^{\frac{1}{2}}}{1} = \sqrt{2}$$
$$x_{2} = \frac{\left(-1\right) 2^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{2}$$
$$x_{3} = \frac{0}{1} + \frac{1 \left(-4\right)^{\frac{1}{2}}}{1} = 2 i$$
$$x_{4} = \frac{0}{1} + \frac{\left(-1\right) \left(-4\right)^{\frac{1}{2}}}{1} = - 2 i$$
$$x^{4} + 2 x^{2} - 8 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} + 2 v - 8 = 0$$
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
$$a = 1$$
$$b = 2$$
$$c = -8$$
, then
D = b^2 - 4 * a * c =
(2)^2 - 4 * (1) * (-8) = 36
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 2$$
Simplify
$$v_{2} = -4$$
Simplify
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:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 2^{\frac{1}{2}}}{1} = \sqrt{2}$$
$$x_{2} = \frac{\left(-1\right) 2^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{2}$$
$$x_{3} = \frac{0}{1} + \frac{1 \left(-4\right)^{\frac{1}{2}}}{1} = 2 i$$
$$x_{4} = \frac{0}{1} + \frac{\left(-1\right) \left(-4\right)^{\frac{1}{2}}}{1} = - 2 i$$
Rapid solution
[src]
___ x1 = -\/ 2
$$x_{1} = - \sqrt{2}$$
___ x2 = \/ 2
$$x_{2} = \sqrt{2}$$
x3 = -2*I
$$x_{3} = - 2 i$$
x4 = 2*I
$$x_{4} = 2 i$$
Sum and product of roots
[src]
sum
___ ___ 0 - \/ 2 + \/ 2 - 2*I + 2*I
$$\left(\left(\left(- \sqrt{2} + 0\right) + \sqrt{2}\right) - 2 i\right) + 2 i$$
=
0
$$0$$
product
___ ___ 1*-\/ 2 *\/ 2 *-2*I*2*I
$$2 i - 2 i \sqrt{2} \cdot 1 \left(- \sqrt{2}\right)$$
=
-8
$$-8$$
-8
The graph