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Quadratic equation
When x = -2,
When x = 3,
The solutions to the equation is -2 or 3. It explains the concept of quadratic equation and how to solve the equation.
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MA105 Homework Assistance (30 questions)
Determine whether the equation is an identity, a conditional equation, or an inconsistent equation...
Solve the equation...
Solve the equation...
Plot the given point in a rectangular coordinate system...
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Linear equations
A linear equation in one variable
? A linear equation in two variables
? A quadratic equation
? A polynomial of three terms
? An exponential function
? A logarithmic function ?
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Solving quadratic/cubic equation
283412 Solving quadratic/cubic equation 57
An equation of the form ax2 + bx + c, where a 0, is called a equation.
Solve the equation.
(x - 1)(x - 3) = 0
x = (smaller value)
x = (larger value)
Solve the equation.
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Solve the differential equation or initial value problem using the method for undetermined constants and variation of parameters.
So the solution is
(2) the particular equation to the differential equation
Assume it has the particular solution has the form of
Plug into the original equation
The particular solution is
(c) the general solution to the original equation
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Equation of a line: Two problems
in above equation we have
-1 = -5*(-4) + c
Gives c = - 21
Substituting in equation (2) we get the equation of the line as
y = -5x - 21 Slope and coordinates of a point are given and the equation of the line are derived.
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Damped harmonic oscillator
get an equation for $lambda$:
begin{equation}
lambda^{2} + 2betalambda +frac{k}{m}=0
end{equation}
We can write the solutions of this equation as:
begin{equation}
lambda_{pm}=-betapm i sqrt{frac{k}{m}-beta^{2}}
end{equation}
This means that the
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Algebra 3rd Set of Problems
Using elimination method, we have:
We'll use either equation to get x: We'll use equation 1:
The answer is letter a.
36. We'll substitute y in equation 1 to y in equation 2. We'll have:
Using either equation, we'll solve for y.
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Algebraically solve for x
We define , the equation is then
So if we multiply B to both sides of the equation, we have
Therefore, the equation is
The right hand of the equation only has constants, so it can be simplified as:
The equation still has two fractions
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equation integers, rational numbers, or irrational numbers
I will claim that x = 1 constitutes a root of that equation ( replace x by 1 in the equation to verify ). Thus, the binomial (x - 1) will divide our cubic equation evenly, without remainder.