# 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.

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### 3 Answers

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.**

kx^2-6x+1=0

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.