Question

Find the value(s) for k for which the roots are real and equal.

(k+1)x² + 2(k+3)x + ( k+8) = 0

Answer 1

(k+1)x² + 2(k+3)x + ( k+8) = 0

roots are real and equal , hence , b² - 4ac = 0

{2(k+3)}² - 4(k+1)(k+8) = 0

4(k+3)² - 4(k² + 9k + 8) = 0

4(k² + 6k + 9) - 4k² - 36k - 32 = 0

4k² + 24k + 36 - 4k² - 36k - 32 = 0

-12k + 4 = 0

4 = 12k

k = 4/12
 
   = 1/3

Answer 2

Hey!!
Given polynomial function is ,say "f(x)"
f(x) = (k+1)x² + 2(k+3)x + ( k+8) = 0

Now roots will be real and equal only when discriminant= 0
We know discriminant is given by :-
d = b²-4ac
where b is coefficient of x ,a is coefficient of x² and c is constant term .
Here b = 2(k+3)
a = (k+1)
c = (k+8)

Now just plug in the values :-
we get
d = { (2k+6)² - 4(k+8)(k+1) }
put d = 0

=> (2k+6)² = 4(k²+9k+8)
=> 4k²+36+24k =4k²+36k + 32
=> -4 = -12k
=> k = 1/3