Question

If the roots of the quadratic equation kx2 + (a + b)x + ab = 0 are (-a, -b).
then the value of k is

Answer 1

Answer:

Step-by-step explanation:

y = a(x - h)^2 + k => where (h, k) is the coordinates of the vertex so in this case:

y = ax^2 + k => the only information given is that the x-coordinate of your vertex(h) is zero thus you need to find a and k, but they did provide 2 points that are on the parabola therefore we use that data to solve for unknowns:

(x, y) = (1, 3) => we can plug in this data:

3 = a * 1^2 + k

=> a + k = 3 => eq-1

(x, y) = (3, 13) => do the same and plug in for x and y:

13 = a * 3^2 + k

=> 9a + k = 13 => eq-2

Now you got the famous 2 equations with 2 unknowns, let’s eliminate k’s by multiplying eq-1 by -1 , then add the 2 equations to eliminate k and solve for a:

-a - k = -3

9a + k = 13 => add the two eq’s:

8a = 10

a = 10/8 = 5/4 => plug in one of the eq’s and solve for k:

5/4 + k = 3

k = 3 - 5/4 = 12/4 - 5/4 = 7/4

Thus: a = 5/4 and k = 7/4 => answer

So the equation becomes:

y = (5/4)x^2 + 7/4

In the link below you can see the graph, notice that the vertex is above the x-axis and the parabola does not cross the x- axis, which means the roots of this quadratic equation are not real and they are complex: