Find the solution to the system
y = 2x2 - 2x - 3
y = -2x + 2
A) (-1,1) and (1, -3)
B) (0, -1) and (-0.5, 0)
C) (0.5, -3.5) and (0, -3)
D) (2, 1) and (0,0)
Can someone please help me I tried doing it and I did not come out with anyone of these answers.
That is what I did but none of those r the answers.
oh, yeah? Too bad you didn't show your work...
substitute #2 into #1 and you have
-2x+2 = 2x^2 - 2x - 3
2x^2 - 5 = 0
x = ±√(5/2)
I suspect a typo, but this is the way to do it.
To find the solution to the system of equations, we need to find the values of x and y that satisfy both equations simultaneously.
Let's set the two given equations equal to each other:
2x^2 - 2x - 3 = -2x + 2
Now, we can simplify this equation by combining like terms:
2x^2 - 2x - 3 + 2x - 2 = 0
Simplifying further gives:
2x^2 - 5 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 2, b = 0, and c = -5. Plugging these values into the quadratic formula:
x = (0 ± √(0^2 - 4*2*(-5))) / (2*2)
x = (0 ± √(0 + 40)) / 4
x = (0 ± √40) / 4
x = (0 ± √(4*10)) / 4
x = (0 ± 2√10) / 4
x = 0 ± √10 / 2
This gives two possible values for x: x = √10 / 2 and x = -√10 / 2.
Now, we can substitute these values of x back into one of the original equations to find the corresponding values of y.
Using the second equation y = -2x + 2:
For x = √10 / 2:
y = -2(√10 / 2) + 2
y = -√10 + 2
For x = -√10 / 2:
y = -2(-√10 / 2) + 2
y = √10 + 2
Therefore, the solutions to the system of equations are:
(√10 / 2, -√10 + 2) and (-√10 / 2, √10 + 2)
None of the given answer choices match the solutions obtained, so it seems there may be a mistake in the options provided. Double-check your work or consult with the source of the problem to get the correct answer.