Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation 2x=18-x
-
Equation x^4=(2*x-3)^2
-
Equation 9*x^2=54*x
-
Equation cos(x)+sin(x)=0
- Express {x} in terms of y where:
- -18*x-1*y=-11
- 7*x-2*y=20
- -1*x-20*y=-17
- 9*x+18*y=3
- Derivative of:
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x^4-5*x^2+4
- Graphing y =:
- x^4-5*x^2+4
- Factor squared:
- x^4-5*x^2+4
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Identical expressions
- x^ four - five *x^ two + four = zero
- x to the power of 4 minus 5 multiply by x squared plus 4 equally 0
- x to the power of four minus five multiply by x to the power of two plus four equally zero
- x4-5*x2+4=0
- x⁴-5*x²+4=0
- x to the power of 4-5*x to the power of 2+4=0
- x^4-5x^2+4=0
- x4-5x2+4=0
- x^4-5*x^2+4=O
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Similar expressions
- x^4-5*x^2-4=0
- x^4+5*x^2+4=0
x^4-5*x^2+4=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\left(x^{4} - 5 x^{2}\right) + 4 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 5 v + 4 = 0$$
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
$$a = 1$$
$$b = -5$$
$$c = 4$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = 4$$
$$v_{2} = 1$$
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:
$$x_{1} = $$
$$\frac{0}{1} + \frac{4^{\frac{1}{2}}}{1} = 2$$
$$x_{2} = $$
$$\frac{\left(-1\right) 4^{\frac{1}{2}}}{1} + \frac{0}{1} = -2$$
$$x_{3} = $$
$$\frac{0}{1} + \frac{1^{\frac{1}{2}}}{1} = 1$$
$$x_{4} = $$
$$\frac{\left(-1\right) 1^{\frac{1}{2}}}{1} + \frac{0}{1} = -1$$
$$\left(x^{4} - 5 x^{2}\right) + 4 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 5 v + 4 = 0$$
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
$$a = 1$$
$$b = -5$$
$$c = 4$$
, then
D = b^2 - 4 * a * c =
(-5)^2 - 4 * (1) * (4) = 9
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 4$$
$$v_{2} = 1$$
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:
$$x_{1} = $$
$$\frac{0}{1} + \frac{4^{\frac{1}{2}}}{1} = 2$$
$$x_{2} = $$
$$\frac{\left(-1\right) 4^{\frac{1}{2}}}{1} + \frac{0}{1} = -2$$
$$x_{3} = $$
$$\frac{0}{1} + \frac{1^{\frac{1}{2}}}{1} = 1$$
$$x_{4} = $$
$$\frac{\left(-1\right) 1^{\frac{1}{2}}}{1} + \frac{0}{1} = -1$$
Rapid solution
[src]
x1 = -2
$$x_{1} = -2$$
x2 = -1
$$x_{2} = -1$$
x3 = 1
$$x_{3} = 1$$
x4 = 2
$$x_{4} = 2$$
x4 = 2
Sum and product of roots
[src]
sum
-2 - 1 + 1 + 2
$$\left(\left(-2 - 1\right) + 1\right) + 2$$
=
0
$$0$$
product
-2*(-1)*2
$$2 \left(- -2\right)$$
=
4
$$4$$
4
The graph