Mister Exam
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How to use it?
- Equation:
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Equation 2tgx–9ctgx+3=0
- Equation (x^2-16):(x^3+3x^2+16)=0
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Equation 16+6x=5(1-2x)-13
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Identical expressions
- (x^ two - sixteen):(x^ three +3x^ two + sixteen)= zero
- (x squared minus 16):(x cubed plus 3x squared plus 16) equally 0
- (x to the power of two minus sixteen):(x to the power of three plus 3x to the power of two plus sixteen) equally zero
- (x2-16):(x3+3x2+16)=0
- x2-16:x3+3x2+16=0
- (x²-16):(x³+3x²+16)=0
- (x to the power of 2-16):(x to the power of 3+3x to the power of 2+16)=0
- x^2-16:x^3+3x^2+16=0
- (x^2-16):(x^3+3x^2+16)=O
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Similar expressions
- (x^2-16):(x^3+3x^2-16)=0
- (x^2-16):(x^3-3x^2+16)=0
- (x^2+16):(x^3+3x^2+16)=0
(x^2-16):(x^3+3x^2+16)=0 equation
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The solution
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2 x - 16 -------------- = 0 3 2 x + 3*x + 16
$$\frac{x^{2} - 16}{\left(x^{3} + 3 x^{2}\right) + 16} = 0$$
Detail solution
Given the equation:
$$\frac{x^{2} - 16}{\left(x^{3} + 3 x^{2}\right) + 16} = 0$$
transform:
Take common factor from the equation
$$\frac{x - 4}{x^{2} - x + 4} = 0$$
the denominator
$$x^{2} - x + 4$$
then
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$x - 4 = 0$$
solve the resulting equation:
1.
$$x - 4 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = 4$$
We get the answer: x1 = 4
but
The final answer:
$$x_{1} = 4$$
$$\frac{x^{2} - 16}{\left(x^{3} + 3 x^{2}\right) + 16} = 0$$
transform:
Take common factor from the equation
$$\frac{x - 4}{x^{2} - x + 4} = 0$$
the denominator
$$x^{2} - x + 4$$
then
x is not equal to 1/2 - sqrt(15)*I/2
x is not equal to 1/2 + sqrt(15)*I/2
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$x - 4 = 0$$
solve the resulting equation:
1.
$$x - 4 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = 4$$
We get the answer: x1 = 4
but
x is not equal to 1/2 - sqrt(15)*I/2
x is not equal to 1/2 + sqrt(15)*I/2
The final answer:
$$x_{1} = 4$$