Use the substitution method to determine whether the linear-quadratic system y=x2−2x−6 , y=−5x−9 has 0, 1, or 2 solutions.(1 point)
To use the substitution method, we start with the second equation, which is y = -5x - 9. We substitute this expression for y into the first equation, y = x^2 - 2x - 6:
-5x - 9 = x^2 - 2x - 6
Rearranging and combining like terms, we get:
x^2 + 3x - 3 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, a = 1, b = 3, and c = -3. Substituting these values into the quadratic formula, we have:
x = (-3 ± sqrt(3^2 - 4(1)(-3))) / (2(1))
x = (-3 ± sqrt(9 + 12)) / 2
x = (-3 ± sqrt(21)) / 2
Since the discriminant (the value inside the square root) is positive (21 > 0), we have two distinct solutions for x.
Therefore, the linear-quadratic system y = x^2 - 2x - 6 and y = -5x - 9 has 2 solutions.