0
k can have any value; however, the range of values permitted depends upon different things:
The value of k depends on the value of x (ie given a value of x, the value of k can be calculated so that kx² + 4x + 5 = 0 has a root at that value of x):
kx² + 4x + 5 = 0
=> kx² = - 4x - 5 = -(4x + 5)
=> k = -(4x + 5)/x²
Note that if x = 0, then the value of k is not determinable.
Another possible answer using the discriminant of b²-4ac; from this the number of roots of the equation can be discovered:
Two real roots:
b²-4ac > 0
→ 4² - 4×k×5 > 0
→ 16 - 20k > 0
→ 20k < 16
→ k < 4/5
So for all values of k less than 4/5 there are two real roots of the quadratic kx² +4x + 5 = 0
One real repeated root:
b² - 4ac = 0
→ k = 4/5
So for k = 4/5, the quadratic (4/5)x² +4x +5 = 0 (→ 4x² +20x + 25 = 0) has one repeated real root.
Two complex roots:
b² - 4ac < 0
→ k > 4/5
So for all values of k greater than 4/5 there are two complex roots of the quadratic kx² +4x + 5 = 0
What else can I help you with?
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