solve the system by the method of substitution
| y= x^3 - x^2 +2
|
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|
| y= x^2 +15x +2
a. (5, 102), (–3, –34), (0, 2)
b. (5, 102), (–1, –12)
c. (5, 102), (3, 56), (1, 18)
d. no real solution
e. (–3, –34), (1, 18)
To find the points of intersection of y = x3 - x2 + 2 and y = x2 + 15x + 2 by substitution:
Set the equations equal to each other: x3 - x2 + 2 = x2 + 15x + 2
Subtract x2 + 15x + 2 from both sides: x3 - x2 + 2 - (x2 + 15x + 2) = 0
Simplify: x3 - x2 + 2 - x2 - 15x - 2 = 0
x3 - 2x2 - 15x = 0
Factor: x(x - 5)(x + 3) = 0
The solutions are: x = 0 and x = 5 and x = -3
Placing these values for x back into y = x2 + 15x + 2 gives these corresponding values for y:
x = 0 ---> y = (0)2 + 15(0) + 2 ---> y = 2 ---> (0, 2)
x = 5 ---> y = (5)2 + 15(5) + 2 ---> y = 102 ---> (5, 102)
x = -3 ---> y = (-3)2 + 15(-3) + 2 ---> y = -34 ---> (-3, -34)
So, (a) is correct.