Determine the quadratic equation whose solutions are: 3 and -2. Explain your solution
if the solutions are x=3 and x=-2
then the quadratic factored to
(x-3)(x+2) = 0
or
x^2 - x - 6 = 0
in general, if a root or solution to some equation is
x = a , then
x - a is a factor of the original equation
If the rootsof aquadratic eqation are -2 and 3detrimne the quadratic equation
Why did the quadratic equation go to the circus? Because it wanted to find its solutions!
To find a quadratic equation with the given solutions of 3 and -2, we can use the fact that the roots of a quadratic equation are equal to the opposite of the coefficients in the equation.
So, the equation can be written as (x - 3)(x + 2) = 0.
Expanding this equation, we get x^2 - x + 6 = 0.
And there you have it, the quadratic equation with solutions 3 and -2 is x^2 - x + 6 = 0.
To determine the quadratic equation whose solutions are given as 3 and -2, we can use the fact that the solutions of a quadratic equation can be found using the formula:
x = (-b ± √(b^2-4ac)) / 2a
In this formula, x represents the solutions, a, b, and c are coefficients of the quadratic equation (ax^2 + bx + c = 0), and ± indicates that there are two possible solutions, one with a plus sign and one with a minus sign.
First, let's assign a variable to the unknown quadratic equation. Let's use "y" as the variable:
y = ax^2 + bx + c
Now, we can substitute the given solutions, 3 and -2, into the equation:
y = a(3)^2 + b(3) + c ... (equation 1)
y = a(-2)^2 + b(-2) + c ... (equation 2)
Expanding these equations:
y = 9a + 3b + c ... (equation 1)
y = 4a - 2b + c ... (equation 2)
As the solutions should satisfy the quadratic equation, plugging in the given solutions should result in both equations equaling zero. So, we can set them equal to zero:
9a + 3b + c = 0 ... (equation 1 set equal to zero)
4a - 2b + c = 0 ... (equation 2 set equal to zero)
Now we have a system of linear equations with three variables (a, b, and c). To solve this system of equations, we can use various methods such as substitution or elimination.
Let's use elimination method to solve this system. We can subtract equation 2 from equation 1:
(9a + 3b + c) - (4a - 2b + c) = 0
After simplification:
9a + 3b - 4a + 2b = 0
5a + 5b = 0
Dividing by 5:
a + b = 0
So, we have found that a + b = 0. Now, we can express one variable in terms of the other and substitute it into either equation 1 or equation 2 to find the value of the remaining variable.
Let's express b in terms of a:
b = -a
Now, substitute this into equation 1:
9a + 3(-a) + c = 0
After simplification:
6a + c = 0
We can express c in terms of a:
c = -6a
Now, we have the values of two variables in terms of a: b = -a and c = -6a.
Finally, we can substitute these values into the quadratic equation in terms of "y":
y = ax^2 + bx + c
y = ax^2 + (-a)x + (-6a)
y = ax^2 - ax - 6a
Now, we can factor out "a" from each term:
y = a(x^2 - x - 6)
So, the quadratic equation whose solutions are 3 and -2 is given by:
y = a(x^2 - x - 6)
Here, "a" can be any non-zero real number, as multiplying the whole equation by a constant will result in a different but equivalent quadratic equation with the same solutions.