Question

for what value of k the quadratic equation (k+4)x²+(k-1)x+1=0 has real and equal roots.​

Answer 1

Answer:

Hi friend!!

➡️We know that, for a given equation ax²+bx+c=0 to have equal roots, the required condition is

b2-4ac=0

➡️Given,

(k+4)x2+k+1)x +1=0 has equal roots

(k+1)²-4(k+4)=0

k2+1+2k-4k-16=0

k2-2k-15=0

k2-5k+3k-15=O

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

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

k=-3,5

If k=-3

Now, the polynomial becomes

(k+4)x2+(k+1)x +1=0

(-3+4)x²+(-3+1)x+1=0

x2-2x+1=0

(x-1)2=0

x=1

If k=5

Now, the polynomial becomes (k+4)x2+(k+1)x +1=0

9x²+6x+1=0

9x?+3x+3x+1=o

3x(3x+1)+1(3x+1)=0

(3x+1)2=0

3x+1=0

x=-1/3

Now, the roots are 1 (or)-1/3

Answer 2

k=5 or k=-3 which has real and equal root.

Open the attachment for detail.