Determine k, k > 0 so that x^2-kx-(k+1)=0 has roots whose difference is 6.?

Answer 1

The roots are (k + or - sqrt(k*k+4(k+1)))/2. So the difference between them is precisely sqrt(k*k+4(k+1)), because of the /2 on the bottom.
So we want k*k+4(k+1) = 6*6 = 36.
k^2 + 4k + 4 = 36
k^2 + 4k - 32 = 0
(k+8)(k-4) = 0
k = 4, since k>0.

A second method would be to notice you can immediately factorise that as (x-(k+1))(x+1). So the roots are k+1 and -1, so k must be 4 to make the difference 6.

Answer 2

This equation will have the form (x - m)(x - n) = 0,
where the roots are m and n.

The difference between the roots is : m - n = 6.

Expanding the equation gives :
x^2 - x(m + n) + mn = 0

Equating coefficients to your equation gives :
m + n = - k
mn = - (k + 1)

From m - n = 6 and m + n = - k, we get :
m = (6 - k) / 2 and n = - (6 + k) / 2

Substituting these into mn = - (k + 1) gives :
[ (6 - k) / 2 ] [ - (6 + k) / 2 ] = -(k + 1)

which gives :
k^2 + 4k - 32 = 0
or :
(k + 8)(k - 4) = 0

Thus, k = -8 or +4.
But you want k > 0, therefore, k = 4

If k = 4, then m = 1 and n = -5,
so your equation becomes :
x^2 - 4x - 5 = 0,
or, (x - 1)(x + 5) = 0

and the difference between the roots is :
m - n = 1 - (-5) = 6

Answer 3

With any k (k>0), the equation always has 2 roots:
_ 1 root is x=-1
_ 1 root is x=k+1
Because:
(x^2 - kx - (k+1) = 0)
<=> (x+1)(x-(k+1)) = 0
<=> x+1=0 or x-(k+1)=0
<=> x=-1 or x=k+1
We have: k>0 <=> k+1>1>-1
So, if the difference between the roots is 6, we have:
k+1 - (-1) = 6 <=> k + 2 = 6 <=> k = 4

Answer 4

Let the roots be a, and a+4, then expand (x-a) ( x - (a+4)) = 0 x² + (-2a-4)x + a²+4a = 0 Equate coefficients with that given in question (-2a-4) = k a²+4a = -(k+1) Solve simultaneously: Therefore k=-6, k=2 Since k<0, then k=-6

Answer 5

x^2 - kx = k+1
x^2 - kx + (k/2)^2 = (k/2)^2 + k + 1 = (1/4)(k^2 + 4k + 4)
x - k/2 = +/- (k+2)/2
Difference in roots = k+2 = 6
k = 4 (Answer)

Check: x^2 -4x - 5 = (x-5)(x+1) = 0

Roots are 5, -1; difference is 6

Answer 6

delta = k^2 + 4(k+1) = (k+2)^2 >=0
it means this equation always has roots
x1 = [k + (k+2)]/2 = k+1
x2 = [k - (k+2)] / 2 = -1
|x1 - x2| = 6 => k = -8 or k = 4
the result is that k=4

Answer 7

x² - kx -(k + 1) = 0

I just did trial and error:
x² - (4)x - ((4) + 1) = 0
x² - 4x - 5 = 0
(x + 1)(x - 5) = 0
Roots are: -1 and 5, where k = 4

Answer 8

k = 4 or -8, but 4 is positive, so it's 4

Answer 9

guess k=4...