Determine if each system has one, two, or no solutions. (Show your work - 3 marks each)
a)y=2x^2-2x+1 and y=3x-5
b)y=x^2+3x-16 and y=-x^2-8x-18
Same thing as your last post
Just evaluate b^2 - 4ac and decide
So I use the quadratic formula?
You could but you don't have to
recall that
x = (-b ± √(b^2 - 4ac)/(2a)
Judging from the type of question you have been asking, I trust you have learned the properties of the "discriminant" b^2 - 4ac
I gave them to you in one of the posts.
To determine the number of solutions for each system, we need to find where the two equations intersect. If they intersect at one point, the system has one solution. If they intersect at multiple points, the system has two solutions. If they do not intersect at any point, the system has no solution.
a) y=2x^2-2x+1 and y=3x-5
To find the solutions, we can set the two equations equal to each other:
2x^2-2x+1 = 3x-5
Rearranging the equation to standard form:
2x^2 - 5x + 6 = 0
To find the solutions, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 2, b = -5, and c = 6. Substituting these values into the quadratic formula, we get:
x = (-(-5) ± sqrt((-5)^2 - 4(2)(6))) / (2(2))
x = (5 ± sqrt(25 - 48)) / 4
x = (5 ± sqrt(-23)) / 4
Since we have a negative value inside the square root, there are no real solutions for x. Therefore, the system has no solution.
b) y=x^2+3x-16 and y=-x^2-8x-18
Setting the equations equal to each other:
x^2 + 3x - 16 = -x^2 - 8x - 18
Rearranging the equation to standard form:
2x^2 + 11x - 2 = 0
Using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 2, b = 11, and c = -2. Substituting these values into the quadratic formula:
x = (-(11) ± sqrt((11)^2 - 4(2)(-2))) / (2(2))
x = (-11 ± sqrt(121 + 16)) / 4
x = (-11 ± sqrt(137)) / 4
Since the value inside the square root is positive, there are two real solutions for x. Therefore, the system has two solutions.
In summary:
a) The system has no solution.
b) The system has two solutions.