Question

find the value of k for which the quadratic equation (k+4)x^2 + (k+1)x + 1 = 0​

Answer 1

Answer:

Step-by-step explanation:

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=0

k(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

9x²+3x+3x+1=0

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

I hope this will help you;)

Answer 2

the value of D of this equation must be greater than or equal to the 0 , after solving equations , k=1 or -1/3