Question

my question is please give my answers

Answer 1

Given Equation is (k + 4)x^2 + (k + 1)x + 1 = 0.

It is in the form of ax^2 + bx + c = 0, Where a = (k + 4), b = (k + 1), c = 1.

Given that the equation has real roots.

D = b^2 - 4ac

= > (k + 1)^2 - 4(k + 4)(1)

= > (k^2 + 1 + 2k - 4k - 16

= > k^2 - 2k - 15

Now,

The equation has real roots.

D = 0

= > k^2 - 2k - 15 = 0

= > k^2 - 5k + 3k - 15 = 0

= > k(k - 5) + 3(k - 5) = 0

= > (k + 3)(k - 5) = 0

= > k - 3 (or) k = 5.



Hope this helps!

Answer 2

Hi,

Please see the attached file!



Thanks