Question

find the value of k for equal roots of (k+1)x2-2(k+4)x+2k=0

Answer 1

(k+4)x2+(k+1)x+1=0
D=b2-4ac
=(k+1)2-4(k+4)(1)
=k2+2k+1-4k-16
=k2-2k-15
For equal roots, D=0
D=0
K2-2k-15=0
k2-5k+3k-15=0
k(k-5)+3(k-5)=0
(k+3)(k-5)=0
k+3=0 OR k-5=0
k=-3 , k=5

Answer 2

here is ur answer dear
a=(k+1) b=-2 (k+4) c=2k
x= -b +- root b^2-4ac/2 a
x= 2k+4 +- root (2k+8)^2 -4 (k+1)(2k)/2 (k+1)
x=2k+4+- root 4k^2 +32k +64 - 8k^2 -8k/2 (k+1)
= 2k+4 +-root -4k^2 +24k +64 /2 (k+1)
=2k+4+-root -4(k+8)^2/2 (k+1)
=2k+4 +-2 (k+8) root -1 / 2 (k+1)
=k+2+-(k+8 ) root -1/k+1
x=k+1+(k+8)root-1/k+1 or k+1-(k+8)root-1/k+1
pls check out the answer