Mister Exam
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How to use it?
- Equation:
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Equation x^2=11
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Equation 10(x-9)=7
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Equation (x+2)^4+(x+2)^2-12=0
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Identical expressions
- (x+ two)^ four +(x+ two)^ two - twelve = zero
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Similar expressions
- (x-2)^4+(x+2)^2-12=0
- (x+2)^4-(x+2)^2-12=0
- (x+2)^4+(x-2)^2-12=0
- (x+2)^4+(x+2)^2+12=0
(x+2)^4+(x+2)^2-12=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
4 2 (x + 2) + (x + 2) - 12 = 0
$$\left(x + 2\right)^{4} + \left(x + 2\right)^{2} - 12 = 0$$
Detail solution
Given the equation:
$$\left(x + 2\right)^{4} + \left(x + 2\right)^{2} - 12 = 0$$
Do replacement
$$v = \left(x + 2\right)^{2}$$
then the equation will be the:
$$v^{2} + v - 12 = 0$$
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
$$a = 1$$
$$b = 1$$
$$c = -12$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = 3$$
Simplify
$$v_{2} = -4$$
Simplify
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:
$$x_{1} = - \frac{2}{1} + \frac{1 \cdot 3^{\frac{1}{2}}}{1} = -2 + \sqrt{3}$$
$$x_{2} = - \frac{2}{1} + \frac{\left(-1\right) 3^{\frac{1}{2}}}{1} = -2 - \sqrt{3}$$
$$x_{3} = - \frac{2}{1} + \frac{1 \left(-4\right)^{\frac{1}{2}}}{1} = -2 + 2 i$$
$$x_{4} = - \frac{2}{1} + \frac{\left(-1\right) \left(-4\right)^{\frac{1}{2}}}{1} = -2 - 2 i$$
$$\left(x + 2\right)^{4} + \left(x + 2\right)^{2} - 12 = 0$$
Do replacement
$$v = \left(x + 2\right)^{2}$$
then the equation will be the:
$$v^{2} + v - 12 = 0$$
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
$$a = 1$$
$$b = 1$$
$$c = -12$$
, then
D = b^2 - 4 * a * c =
(1)^2 - 4 * (1) * (-12) = 49
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 3$$
Simplify
$$v_{2} = -4$$
Simplify
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:
$$x_{1} = - \frac{2}{1} + \frac{1 \cdot 3^{\frac{1}{2}}}{1} = -2 + \sqrt{3}$$
$$x_{2} = - \frac{2}{1} + \frac{\left(-1\right) 3^{\frac{1}{2}}}{1} = -2 - \sqrt{3}$$
$$x_{3} = - \frac{2}{1} + \frac{1 \left(-4\right)^{\frac{1}{2}}}{1} = -2 + 2 i$$
$$x_{4} = - \frac{2}{1} + \frac{\left(-1\right) \left(-4\right)^{\frac{1}{2}}}{1} = -2 - 2 i$$
Rapid solution
[src]
___ x1 = -2 - \/ 3
$$x_{1} = -2 - \sqrt{3}$$
___ x2 = -2 + \/ 3
$$x_{2} = -2 + \sqrt{3}$$
x3 = -2 - 2*I
$$x_{3} = -2 - 2 i$$
x4 = -2 + 2*I
$$x_{4} = -2 + 2 i$$
Sum and product of roots
[src]
sum
___ ___ 0 + -2 - \/ 3 + -2 + \/ 3 + -2 - 2*I + -2 + 2*I
$$\left(\left(\left(\left(-2 - \sqrt{3}\right) + 0\right) - \left(2 - \sqrt{3}\right)\right) - \left(2 + 2 i\right)\right) - \left(2 - 2 i\right)$$
=
-8
$$-8$$
product
/ ___\ / ___\ 1*\-2 - \/ 3 /*\-2 + \/ 3 /*(-2 - 2*I)*(-2 + 2*I)
$$1 \left(-2 - \sqrt{3}\right) \left(-2 + \sqrt{3}\right) \left(-2 - 2 i\right) \left(-2 + 2 i\right)$$
=
8
$$8$$
8
Numerical answer
[src]
x1 = -2.0 + 2.0*i
x2 = -0.267949192431123
x3 = -2.0 - 2.0*i
x4 = -3.73205080756888
x4 = -3.73205080756888
The graph