Mister Exam
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How to use it?
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Similar expressions
- x^4-x^2+20=0
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x^4-x^2-20=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\left(x^{4} - x^{2}\right) - 20 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - v - 20 = 0$$
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
$$a = 1$$
$$b = -1$$
$$c = -20$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = 5$$
$$v_{2} = -4$$
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:
$$x_{1} = $$
$$\frac{0}{1} + \frac{5^{\frac{1}{2}}}{1} = \sqrt{5}$$
$$x_{2} = $$
$$\frac{\left(-1\right) 5^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{5}$$
$$x_{3} = $$
$$\frac{0}{1} + \frac{\left(-4\right)^{\frac{1}{2}}}{1} = 2 i$$
$$x_{4} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(-4\right)^{\frac{1}{2}}}{1} = - 2 i$$
$$\left(x^{4} - x^{2}\right) - 20 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - v - 20 = 0$$
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
$$a = 1$$
$$b = -1$$
$$c = -20$$
, then
D = b^2 - 4 * a * c =
(-1)^2 - 4 * (1) * (-20) = 81
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 5$$
$$v_{2} = -4$$
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:
$$x_{1} = $$
$$\frac{0}{1} + \frac{5^{\frac{1}{2}}}{1} = \sqrt{5}$$
$$x_{2} = $$
$$\frac{\left(-1\right) 5^{\frac{1}{2}}}{1} + \frac{0}{1} = - \sqrt{5}$$
$$x_{3} = $$
$$\frac{0}{1} + \frac{\left(-4\right)^{\frac{1}{2}}}{1} = 2 i$$
$$x_{4} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(-4\right)^{\frac{1}{2}}}{1} = - 2 i$$
Rapid solution
[src]
___ x1 = -\/ 5
$$x_{1} = - \sqrt{5}$$
___ x2 = \/ 5
$$x_{2} = \sqrt{5}$$
x3 = -2*I
$$x_{3} = - 2 i$$
x4 = 2*I
$$x_{4} = 2 i$$
x4 = 2*i
Sum and product of roots
[src]
sum
___ ___ - \/ 5 + \/ 5 - 2*I + 2*I
$$\left(\left(- \sqrt{5} + \sqrt{5}\right) - 2 i\right) + 2 i$$
=
0
$$0$$
product
___ ___ -\/ 5 *\/ 5 *-2*I*2*I
$$2 i - 2 i - \sqrt{5} \sqrt{5}$$
=
-20
$$-20$$
-20
Numerical answer
[src]
x1 = -2.23606797749979
x2 = -2.0*i
x3 = 2.0*i
x4 = 2.23606797749979
x4 = 2.23606797749979
The graph