Determine the value of k in y=kx^2-5x+2 that will result in the intersection of the line y=-3x+4 with the quadratic at
a) two points (1 mark)
b) one points (1 mark)
c) no point (1 mark)
Here's my work on it:
kx^2-5x+2=-3x+4
kx^2-2x-2=0
a)b^2-4ac
(-2^2-4(1)(-2)
(-4--8)
(4)
Can you please help me with b & c and tell me where I went wrong with a) if I did so. Thanks!
the discriminant is
25 - 8k
so, for one (repeated) root, 25-8k=0
find that, and then any value of k greater will produce a negative discriminant
any lesser value of k will produce a positive discriminant.
To find the values of k that will result in the given intersections, we need to solve the quadratic equation kx^2 - 2x - 2 = 0. Now let's analyze each case:
a) For two intersecting points:
In this case, the discriminant (b^2 - 4ac) of the quadratic equation should be positive. So, you correctly calculated the discriminant:
b^2 - 4ac = (-2)^2 - 4(1)(-2) = 4 + 8 = 12
Since the discriminant is positive, there will be two distinct points of intersection. Therefore, your calculation for case (a) is correct.
b) For one point of intersection:
In this case, the discriminant should be equal to zero. Set the discriminant equal to zero and solve for k:
b^2 - 4ac = 0
(-2)^2 - 4(1)(-2) = 0
4 + 8 = 0
12 = 0 (This is not true, so there is no value of k that will result in just one point of intersection)
Thus, there is no value of k that will result in a single point of intersection.
c) For no point of intersection:
In this case, the discriminant should be negative. Let's try setting k = 0 and check if it satisfies the condition:
b^2 - 4ac = (-2)^2 - 4(0)(-2) = 0 - 0 = 0
Since the discriminant is not negative, k = 0 does not yield no points of intersection.
Therefore, there is no value of k that will result in no point of intersection.
In summary:
a) For two points of intersection, any real value of k will satisfy.
b) There is no value of k that will result in only one point of intersection.
c) There is no value of k that will result in no point of intersection.