Determine the value(s)of k that give the type of solution indicated:
2x^2+5x-2(k-1)=0. Show your work.
To find the value(s) of k that give the type of solution indicated, we need to consider the discriminant of the quadratic equation.
The discriminant is given by the formula: D = b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation in the form of ax^2 + bx + c = 0.
In this case, the given quadratic equation is 2x^2 + 5x - 2(k-1) = 0. Comparing this to the general form ax^2 + bx + c = 0, we can see that a = 2, b = 5, and c = -2(k-1).
Now, let's substitute these values into the discriminant formula: D = (5)^2 - 4(2)(-2(k-1))
Simplifying further, we have: D = 25 + 16(k-1)
To determine the type of solutions, we need to consider the value of the discriminant:
1. If D > 0, then the quadratic equation has two distinct real solutions.
2. If D = 0, then the quadratic equation has one real solution (repeated root).
3. If D < 0, then the quadratic equation has no real solutions (two complex solutions).
Let's analyze these cases:
Case 1: D > 0
For a quadratic equation to have two distinct real solutions, the discriminant must be greater than zero.
25 + 16(k-1) > 0
To solve for k, we need to isolate it:
16(k-1) > -25
k - 1 > -25/16
k > -25/16 + 1
k > -25/16 + 16/16
k > -9/16
Therefore, for D > 0, k must be greater than -9/16.
Case 2: D = 0
For a quadratic equation to have one real solution (repeated root), the discriminant must be equal to zero.
25 + 16(k-1) = 0
16(k-1) = -25
k - 1 = -25/16
k = -25/16 + 16/16
k = -25/16 + 1
k = -9/16
Therefore, for D = 0, k must be equal to -9/16.
Case 3: D < 0
For a quadratic equation to have no real solutions (two complex solutions), the discriminant must be less than zero.
25 + 16(k-1) < 0
16(k-1) < -25
k - 1 < -25/16
k < -25/16 + 16/16
k < -25/16 + 1
k < -9/16
Therefore, for D < 0, k must be less than -9/16.
To summarize,
- If k > -9/16, the quadratic equation has two distinct real solutions.
- If k = -9/16, the quadratic equation has one real solution (repeated root).
- If k < -9/16, the quadratic equation has no real solutions (two complex solutions).
So, the value(s) of k that give the type of solution indicated are k > -9/16.