find the value of k in each of the following?a.) kx^2-6x+1=0 if the equation has real, rational, and equal roots.
a.) kx^2-6x+1=0 if the equation has real, rational, and equal roots.
If the function has rational , real and equal roots, then that mean delta will be 0:
We know that:
delta = b^2 - 4ac
==> delta = (-6)^2 - 4*k*1
==> 36 - 4k = 0
==> -4k = -36
==> k = -36/-4 = 9
==> k = 9
b.) 2x^2+6x+k=0 find the smallest integral value of k, such that the equation had imaginary roots
If the function has imaginary of (complex) roots, then delta should be less than 0 ( negative).
==> delta = b^2 - 4ac
==> delta = (6^2) - 4*2*k < 0
==> 36 - 8k < 0
==> -8k < -36
==> k > -36/-8
==> k > 9/2
Then the smallest integer is the first integer that comes after 9/2
9/2 = 4.5 , then the next integer s 5
Then k = 5 is the smallest integer so the equation has complex roots.
If the equation has real, rational, and equal roots then the discriminant has to be zero.
We'll calculate the discriminant delta:
delta = b^2 - 4ac
We'll identify a,b,c:
a = k, b = -6, c = 1
delta = 36 - 4k
We'll cancel delta:
36 - 4k = 0
We'll isolate k to the left side. For this reason, we'll subtract 36 both sides:
-4k = -36
We'll divide by -4:
k = 9
So, the roots of the equation are real and equal if and only if k = 9.
The roots are:
x1 = x2 = (-b+sqrt delta)/2a
We'll substitute delta and b:
x1 = x2 = (6 + 0)/2*9
x1 = x2 = 1/3
To find the value of k in kx^2-6x+1=0.
Given that the roots are real rational and equal roots.
Let each root is equal to x1 then ,
sum of the roots = 2x1 = -(-6)/k.
Or x1 = 6/2k = 3/k...(1)
Product of the roots x1^2 = 1/k...(2)
Eliminate x1 from (1) and (2):
From (1): x1^2 = 3^2/(k)^2 = 9/k^2...(3)
From (2) x^2 = 1/k....(4).
From (3) and (4):
9/k^2 = 1/k. Multiply by k^2.
9 = k.
Therefor k = 9.
Therefore for k = 9, the given equation can have equal roots if k = 9.