Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation x^4-2x^2-8=0
-
Equation tgx=-(3/4)
-
Equation 7(4x-1)=6-2(3-14x)
- Equation 836-16x=420
- Express {x} in terms of y where:
- (x^2+y^2-1)^3+x^2*y^3=0
- -1*x-19*y=-10
- 4*x-16*y=6
- 5*x+5*y=20
- Graphing y =:
- x^4-2x^2-8
-
Identical expressions
- x^ four - two x^2- eight = zero
- x to the power of 4 minus 2x squared minus 8 equally 0
- x to the power of four minus two x squared minus eight equally zero
- x4-2x2-8=0
- x⁴-2x²-8=0
- x to the power of 4-2x to the power of 2-8=0
- x^4-2x^2-8=O
-
Similar expressions
- x^4+2x^2-8=0
- x^4-2x^2+8=0
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
$$a\ v^2 + b\ v + c = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = -2$$
$$c = -8$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-2\right)^{2} - 1 \cdot 4 \left(-8\right) = 36$$
Because D > 0, then the equation has two roots.
$$v_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$v_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$v_{1} = 4$$
Simplify
$$v_{2} = -2$$
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:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 4^{\frac{1}{2}}}{1} = 2$$
$$x_{2} = \frac{\left(-1\right) 4^{\frac{1}{2}}}{1} + \frac{0}{1} = -2$$
$$x_{3} = \frac{0}{1} + \frac{1 \left(-2\right)^{\frac{1}{2}}}{1} = \sqrt{2} i$$
$$x_{4} = \frac{0}{1} + \frac{\left(-1\right) \left(-2\right)^{\frac{1}{2}}}{1} = - \sqrt{2} i$$
$$x^{4} - 2 x^{2} - 8 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = -2$$
$$c = -8$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-2\right)^{2} - 1 \cdot 4 \left(-8\right) = 36$$
Because D > 0, then the equation has two roots.
$$v_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$v_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$v_{1} = 4$$
Simplify
$$v_{2} = -2$$
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:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 4^{\frac{1}{2}}}{1} = 2$$
$$x_{2} = \frac{\left(-1\right) 4^{\frac{1}{2}}}{1} + \frac{0}{1} = -2$$
$$x_{3} = \frac{0}{1} + \frac{1 \left(-2\right)^{\frac{1}{2}}}{1} = \sqrt{2} i$$
$$x_{4} = \frac{0}{1} + \frac{\left(-1\right) \left(-2\right)^{\frac{1}{2}}}{1} = - \sqrt{2} i$$
Sum and product of roots
[src]
sum
___ ___ -2 + 2 + -I*\/ 2 + I*\/ 2
$$\left(-2\right) + \left(2\right) + \left(- \sqrt{2} i\right) + \left(\sqrt{2} i\right)$$
=
0
$$0$$
product
___ ___ -2 * 2 * -I*\/ 2 * I*\/ 2
$$\left(-2\right) * \left(2\right) * \left(- \sqrt{2} i\right) * \left(\sqrt{2} i\right)$$
=
-8
$$-8$$
Rapid solution
[src]
x_1 = -2
$$x_{1} = -2$$
x_2 = 2
$$x_{2} = 2$$
___ x_3 = -I*\/ 2
$$x_{3} = - \sqrt{2} i$$
___ x_4 = I*\/ 2
$$x_{4} = \sqrt{2} i$$
The graph