How many different points of intersection are there for x + y = 7 and y = x squared minus 7?
equate
x^2 - 7 = 7-x
x^2 + x - 14 = 0
this equation has two different real roots
so there are two different points of intersection.
To find the points of intersection between two equations, we can set them equal to each other and solve for the variables. In this case, we want to find the intersection points of the equations:
1. x + y = 7
2. y = x^2 - 7
To get started, let's substitute equation 2 into equation 1:
x + (x^2 - 7) = 7
Now, we have a quadratic equation. Rearrange the equation to put it in standard form:
x^2 + x - 14 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. Since it doesn't factor easily, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = 1, and c = -14. Substituting these values into the formula:
x = (-1 ± √(1^2 - 4(1)(-14))) / (2(1))
Simplifying further:
x = (-1 ± √(1 + 56)) / 2
x = (-1 ± √57) / 2
Now we have two possible values for x. Let's calculate the corresponding y-values by substituting these x-values into equation 2:
For x = (-1 + √57) / 2:
y = ((-1 + √57) / 2)^2 - 7
And for x = (-1 - √57) / 2:
y = ((-1 - √57) / 2)^2 - 7
By calculating these expressions, we can determine the y-values, which correspond to the different points of intersection. So, the answer to the question "How many different points of intersection are there?" will depend on the solutions of these equations.