Question

3. Find the value of k for which the given equation has real and equal roots:
x²+k(4x + k-1) + 2 = 0.​

Answer 1

Answer:

k=1or2/3

x2+k(4x+k-1)+2

=x2+4kx+(k2-k+2)

when the roots are equal be=4ac

so. (4k)2=4*1*(k2-k+2)

16k2=4k2-4k+8

12k2+4k-8=0

3k2+k-2=0

(3k-2)(k+1)=0

k=2/3or-1

so.the polynomial is x2-4x+4=(x-2)2orx2+8x/3+16/9=(x+4/3)2