which of the following sets gives the x coordinates where the parabola y=x^2+4X-50 and the line y=7x-10 intersect when drawn in the coordinate plane
7x-10 = x^2 + 4 x - 50
x^2 -3x -40 = 0
(x-8)(x+5) = 0
x = 8 or x = -5
(-5,-45)
(8,46)
Find the coordinates of the point of intersection of the straight line y=3x_6 with the parabola y=x^2_6x+8
To find the x-coordinates where the parabola and the line intersect, we need to set the equations equal to each other and solve for the values of x.
First, let's set the two equations equal to each other:
x^2 + 4x - 50 = 7x - 10
Now, let's bring all the terms to one side of the equation:
x^2 + 4x - 7x - 50 + 10 = 0
Simplifying, we get:
x^2 - 3x - 40 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. Let's use the quadratic formula since the equation cannot be easily factored:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -3, and c = -40. Plugging in these values, we can calculate the x-coordinates:
x = (3 ± √((-3)^2 - 4(1)(-40))) / (2(1))
Simplifying further:
x = (3 ± √(9 + 160)) / 2
x = (3 ± √169) / 2
x = (3 ± 13) / 2
This gives two possible x-values:
1. x = (3 + 13) / 2 = 16 / 2 = 8
2. x = (3 - 13) / 2 = -10 / 2 = -5
Therefore, the two x-coordinates where the parabola y = x^2 + 4x - 50 and the line y = 7x - 10 intersect are x = 8 and x = -5.
So, the answer would be the set {8, -5}.