Solve the equation y^2+3y+2 = 0 how is the equation y^2+3yx+2x^2 related
how are they differnt
y^2 + 3y + 2 = 0
(y+1)(y+2) = 0
y = -2, -1
y^2 + 3xy + 2x^2 = 0
(y+x)(y+2x) = 0
y = -x, y = -2x
so, what do you think?
How did you solve y^2 + 3xy + 2x^2 = 0? Im confused at the one.
Huh? I just factored it. The solution is two straight lines; it is a degenerate hyperbola.
I mean in Quadratic forms I don't even know what is a degenrate hyperbola.
This is on my test tommrow.
well, a quadratic form as all terms of the 2nd degree. So, the first equation is not even a quadratic form. It is just a quadratic equation.
There are lots of articles and videos available online (google is your friend), so whatever level of study you need is surely accessible there. First place to look -- your class textbook and your other homework assignments. You gotta assume that nothing will be asked on the test which you have not covered in class.
Alright I understood how to do it, i got Confusued cause of the xy
To solve the equation y^2 + 3y + 2 = 0, you can use the quadratic formula. The quadratic formula is defined as:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, we need to make some adjustments to match the equation format.
Let's consider y as our variable and rewrite the equation in the form of ay^2 + by + c = 0. Then, comparing it with the given equation y^2 + 3y + 2 = 0, we get a = 1, b = 3, and c = 2.
Now, we can substitute these values into the quadratic formula:
y = (-3 ± √(3^2 - 4(1)(2))) / (2(1))
y = (-3 ± √(9 - 8)) / 2
y = (-3 ± √1) / 2
Simplifying further:
y = (-3 ± 1) / 2
Now, breaking it down into two solutions:
y = (-3 + 1) / 2 , y = (-3 - 1) / 2
y = -2/2 , y = -4/2
y = -1 , y = -2
Therefore, the solutions to the equation y^2 + 3y + 2 = 0 are y = -1 and y = -2.
Regarding the second part of your question, the equation y^2 + 3yx + 2x^2 is related to the initial equation y^2 + 3y + 2 = 0 through a substitution. In this equation, there are two variables, y and x. By substituting y with x in y^2 + 3y + 2 = 0, you get y^2 + 3yx + 2x^2. However, it's important to note that these two equations have different solutions, as they involve different variables.