Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation (x+6)^3=25*(x+6)
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Equation 1-5x=-6x+8
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Equation 1-5*x=-6*x+8
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Equation x^2-8*x+17=0
- Express {x} in terms of y where:
- -6*x-20*y=8
- -9*x-15*y=-4
- 12*x-14*y=20
- 17*x-20*y=6
- Canonical form:
- (x+6)^3
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Identical expressions
- (x+ six)^ three = twenty-five *(x+ six)
- (x plus 6) cubed equally 25 multiply by (x plus 6)
- (x plus six) to the power of three equally twenty minus five multiply by (x plus six)
- (x+6)3=25*(x+6)
- x+63=25*x+6
- (x+6)³=25*(x+6)
- (x+6) to the power of 3=25*(x+6)
- (x+6)^3=25(x+6)
- (x+6)3=25(x+6)
- x+63=25x+6
- x+6^3=25x+6
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Similar expressions
- cos(10*x+12)+4*sqrt2*(5*x+6)=4
- (x+6)^3=25*(x-6)
- (x-6)^3=25*(x+6)
- ((x+2)^2+2):((x^2)-1+((25x+6):x^4))=0
- (x-6)3=25(x+6)
- (x+6)^3=25(x+6)
(x+6)^3=25*(x+6) equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\left(x + 6\right)^{3} = 25 \left(x + 6\right)$$
transform:
Take common factor from the equation
$$\left(x + 1\right) \left(x + 6\right) \left(x + 11\right) = 0$$
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
$$x + 1 = 0$$
$$x + 6 = 0$$
$$x + 11 = 0$$
solve the resulting equation:
1.
$$x + 1 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = -1$$
We get the answer: x1 = -1
2.
$$x + 6 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = -6$$
We get the answer: x2 = -6
3.
$$x + 11 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = -11$$
We get the answer: x3 = -11
The final answer:
$$x_{1} = -1$$
$$x_{2} = -6$$
$$x_{3} = -11$$
$$\left(x + 6\right)^{3} = 25 \left(x + 6\right)$$
transform:
Take common factor from the equation
$$\left(x + 1\right) \left(x + 6\right) \left(x + 11\right) = 0$$
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
$$x + 1 = 0$$
$$x + 6 = 0$$
$$x + 11 = 0$$
solve the resulting equation:
1.
$$x + 1 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = -1$$
We get the answer: x1 = -1
2.
$$x + 6 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = -6$$
We get the answer: x2 = -6
3.
$$x + 11 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = -11$$
We get the answer: x3 = -11
The final answer:
$$x_{1} = -1$$
$$x_{2} = -6$$
$$x_{3} = -11$$
Sum and product of roots
[src]
sum
-11 - 6 - 1
$$\left(-11 - 6\right) - 1$$
=
-18
$$-18$$
product
-11*(-6)*(-1)
$$\left(-1\right) \left(- -66\right)$$
=
-66
$$-66$$
-66
The graph