Question 1081227: For the quadratic relations y=x²+5x-7 an y=ax²+(a+4)x+9.
a)Determine the values(s) of a, where there is no points of intersection between the quadratic relations.
B)Determine the values(s) of a, where there is one point of intersection between the quadratic relations
C)Determine the values(s) of a, where there is 2 points of intersection between the quadratic relations
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! We have two equations and two unknowns:
y = x^2 + 5x - 7
y = ax^2 + (a+4)x + 9
Setting the y values equal, we obtain a quadratic equation in x:
x^2 + 5x - 7 = ax^2 + (a+4)x + 9 -> (1-a)x^2 + (1-a)x - 16 = 0 [1]
The number of solutions of a quadratic depends on the value of the discriminant, b^2 - 4ac
If b^2 - 4ac < 0 there are no solutions
If b^2 - 4ac = 0 there is one solution
If b^2 - 4ac > 0 there are two solutions
For simplicity, set k = 1-a
kx^2 + kx - 16 = 0
b^2 - 4ac = k^2 + 64k
a) k^2 + 64k < 0
This will be true if k<0 and k>-64 -> 1-a<0 -> a>1 AND 1-a>-64 -> a<65
b) k^2 + 64k = 0
k(k+64) = 0
This has two solutions, k=0 and k=-64, -> a=1 and a=65
But a=1 cannot yield a solution, since from [1], we would have -16 = 0, which is obviously false.
Therefore, a=65 is the only value for a which yields one intersection point
c) can be solved in a similar manner
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