1.

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

Answer» The given equation is `(k+4)x^(2)+(k+1)x+1=0.`
This is of the form `ax^(2)+bx+c=0,` where `a=(k+4), b=(k+1)" and "c=1.`
`:." "D=(b^(2)-4ac)=(k+1)^(2)-4xx(k+4)xx1=(k+1)^(2)-4(k+4)`
`=(k^(2)+1+2k-4k-16)=(k^(2)-2k-15).`
For equal roots, we must have
`D=0impliesk^(2)-2k-15=0`
`implies" "k^(2)-5k+3k-15=0impliesk(k-5)+3(k-5)=0`
`implies" "(k-5)(k+3)=0impliesk-5=0" or "k+3=0`
`implies" "k=5" or "k=-3.`
Hence, the required value of k is or -3.


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