Question

find the value of k for which the quadratic equation has *real roots* (k+4)x2+(k+1)x+1=0

Answer 1

Answer:

k=0 or -7/4

Step-by-step explanation:

As it is a quadratic equation,

we can use the sum of roots and product of roots.

And, It is also mentioned that it is a real root.So, the roots are same.

Let's consider the root as x

Sum of roots=x+x=-(k+1)/(k+4)

                    x=2*-(k+1)/(k+4) ----(1)

Product of roots=x*x=1/(k+4)

                            x^2=1/(k+4)

                            x=(1/(k+4))^1/2   ----(2)

(1)=(2)

-2*(k+1)/(k+4)=(1/(k+4))^1/2

Squaring on both sides,

So,4*(k+1)^2/(k+4)=1

4k^2+8k+4=k+4

4k^2+7k-0=0

4k^2+7k=0

K(4k+7)=0

k=0 or k=-7/4