Mister Exam
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How to use it?
- Equation:
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Equation (x+2)^4-4*(x+2)^2-5=0
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Equation x-6/2-x/3=3
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Equation x^3+2x^2-x-2=0
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Identical expressions
- (x+ two)^ four - four *(x+ two)^ two - five = zero
- (x plus 2) to the power of 4 minus 4 multiply by (x plus 2) squared minus 5 equally 0
- (x plus two) to the power of four minus four multiply by (x plus two) to the power of two minus five equally zero
- (x+2)4-4*(x+2)2-5=0
- x+24-4*x+22-5=0
- (x+2)⁴-4*(x+2)²-5=0
- (x+2) to the power of 4-4*(x+2) to the power of 2-5=0
- (x+2)^4-4(x+2)^2-5=0
- (x+2)4-4(x+2)2-5=0
- x+24-4x+22-5=0
- x+2^4-4x+2^2-5=0
- (x+2)^4-4*(x+2)^2-5=O
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Similar expressions
- (x-2)^4-4*(x+2)^2-5=0
- (x+2)^4+4*(x+2)^2-5=0
- (x+2)^4-4*(x+2)^2+5=0
- (x+2)^4-4*(x-2)^2-5=0
(x+2)^4-4*(x+2)^2-5=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
4 2 (x + 2) - 4*(x + 2) - 5 = 0
$$\left(\left(x + 2\right)^{4} - 4 \left(x + 2\right)^{2}\right) - 5 = 0$$
Detail solution
Given the equation:
$$\left(\left(x + 2\right)^{4} - 4 \left(x + 2\right)^{2}\right) - 5 = 0$$
Do replacement
$$v = \left(x + 2\right)^{2}$$
then the equation will be the:
$$v^{2} - 4 v - 5 = 0$$
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
$$a = 1$$
$$b = -4$$
$$c = -5$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = 5$$
$$v_{2} = -1$$
The final answer:
Because
$$v = \left(x + 2\right)^{2}$$
then
$$x_{1} = \sqrt{v_{1}} - 2$$
$$x_{2} = - \sqrt{v_{1}} - 2$$
$$x_{3} = \sqrt{v_{2}} - 2$$
$$x_{4} = - \sqrt{v_{2}} - 2$$
then:
$$x_{1} = $$
$$- \frac{2}{1} + \frac{5^{\frac{1}{2}}}{1} = -2 + \sqrt{5}$$
$$x_{2} = $$
$$\frac{\left(-1\right) 5^{\frac{1}{2}}}{1} - \frac{2}{1} = - \sqrt{5} - 2$$
$$x_{3} = $$
$$- \frac{2}{1} + \frac{\left(-1\right)^{\frac{1}{2}}}{1} = -2 + i$$
$$x_{4} = $$
$$- \frac{2}{1} + \frac{\left(-1\right) \left(-1\right)^{\frac{1}{2}}}{1} = -2 - i$$
$$\left(\left(x + 2\right)^{4} - 4 \left(x + 2\right)^{2}\right) - 5 = 0$$
Do replacement
$$v = \left(x + 2\right)^{2}$$
then the equation will be the:
$$v^{2} - 4 v - 5 = 0$$
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
$$a = 1$$
$$b = -4$$
$$c = -5$$
, then
D = b^2 - 4 * a * c =
(-4)^2 - 4 * (1) * (-5) = 36
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 5$$
$$v_{2} = -1$$
The final answer:
Because
$$v = \left(x + 2\right)^{2}$$
then
$$x_{1} = \sqrt{v_{1}} - 2$$
$$x_{2} = - \sqrt{v_{1}} - 2$$
$$x_{3} = \sqrt{v_{2}} - 2$$
$$x_{4} = - \sqrt{v_{2}} - 2$$
then:
$$x_{1} = $$
$$- \frac{2}{1} + \frac{5^{\frac{1}{2}}}{1} = -2 + \sqrt{5}$$
$$x_{2} = $$
$$\frac{\left(-1\right) 5^{\frac{1}{2}}}{1} - \frac{2}{1} = - \sqrt{5} - 2$$
$$x_{3} = $$
$$- \frac{2}{1} + \frac{\left(-1\right)^{\frac{1}{2}}}{1} = -2 + i$$
$$x_{4} = $$
$$- \frac{2}{1} + \frac{\left(-1\right) \left(-1\right)^{\frac{1}{2}}}{1} = -2 - i$$
Rapid solution
[src]
___ x1 = -2 + \/ 5
$$x_{1} = -2 + \sqrt{5}$$
___ x2 = -2 - \/ 5
$$x_{2} = - \sqrt{5} - 2$$
x3 = -2 - I
$$x_{3} = -2 - i$$
x4 = -2 + I
$$x_{4} = -2 + i$$
x4 = -2 + i
Sum and product of roots
[src]
sum
___ ___ -2 + \/ 5 + -2 - \/ 5 + -2 - I + -2 + I
$$\left(\left(\left(- \sqrt{5} - 2\right) + \left(-2 + \sqrt{5}\right)\right) + \left(-2 - i\right)\right) + \left(-2 + i\right)$$
=
-8
$$-8$$
product
/ ___\ / ___\ \-2 + \/ 5 /*\-2 - \/ 5 /*(-2 - I)*(-2 + I)
$$\left(-2 + \sqrt{5}\right) \left(- \sqrt{5} - 2\right) \left(-2 - i\right) \left(-2 + i\right)$$
=
-5
$$-5$$
-5
Numerical answer
[src]
x1 = -2.0 - 1.0*i
x2 = 0.23606797749979
x3 = -2.0 + 1.0*i
x4 = -4.23606797749979
x4 = -4.23606797749979
The graph