Question

for what value of k , the equation (k+ 3)x² - (5-k)
x+ 1 = 0 has
(il coincident xonte (iidistinct roots ?​

Answer 1

The given quadratic equation is;

(k+3)x^2 - (5-k)x + 1 = 0.

Then ,

Determinant = b^2 - 4•a•c

=> D = [-(5-k)]^2 - 4•(k+3)•1

=> D = 25 - 10k + k^2 - 4k - 12

=> D = k^2 - 14k + 13

=> D = k^2 - 13k - k + 13

=> D = k(k - 13) - (k - 13)

=> D = (k - 13)(k - 1)

Also,

For the roots to be real and coincident (equal roots) , the determinant of the quadratic equation must be equal to zero.

Thus;

=> D = 0

=> (k - 13)(k - 1) = 0

=> k = 13 or 1.

Hence,

For coincident roots , k and take values of 1 and 13.

Now,

For distinct roots, the determinant must be greater than zero.

Thus;

=> D > 0

=> (k - 13)(k - 1) > 0

Here,

The following two cases are possible.

Case(1):

When , (k - 13) > 0 and (k - 1) > 0

Then , k > 13 and k > 1

=> k > 13

Or

Case(2):

When , (k - 13) < 0 and (k - 1) < 0

Then , k < 13 and k < 1

=> k < 1

Thus,

From cases (1) and (2) , we get;

k is either greater than 13 or less than 1.

Hence,

For the distinct roots, k can take any real value which is greater than 13 or less than 1.

Answer 2

Answer:

k>13 or k<1 this is the correct answer for the given question