Find the value(s) of k for which the given quadratic equations has real and distinct roots :
(i) `2x^(2)+kx+4=0` (ii) `4x^(2)-3kx+1=0`
(iii) `kx^(2)+6x+1=0` (iv) `x^(2)-kx+9=0`

Explanation

(i) The given equation is
`2x^(2)+kx+4=0`
Comparing with `ax^(2)+bx+c=0` we get
a=2, b=k and c=4
`:.D=b^(2)-4acgt=0` for real and distinct roots.
Therefore, `D=k^(2)-4xx2xx(4)gt=0`
`impliesk^(2)-32gt=0`
`impliesk^(2)gt=32`
`impliesklt=-4sqrt2andkgt=4sqrt2`
(ii) The given equation is `4x^(2)-3kx+1=0`
Comparing with `ax^(2)+bx+c=0`, we get
a=4, b=-3k and c=1
`:.D=b^(2)-4acgt=0` for real distinct roots
Therefore, `D=(-3k)^(2)-4xx4xx1gt=0`
`implies9x^(2)-16gt=0implies9k^(2)gt=16`
`impliesk(2)gt=(16)/(9)`
`impliesklt=-(4)/(3)andkgt=(4)/(3)`
(iii) The given equation is
`kx^(2)+6x+1=0`
Comparing with `ax^(2)+bx+c=0`, we get
a=k, b=6 and c=1
`:.D=b^(2)-4acgt=0` for real and distinct roots
Therefore, `D=(6)^(2)-4xxkxx1gt=0`
`impliesk^(2)-36-4kgt=0" " implies" " 36gt=4k`
`impliesklt=9`
(iv) The gievn equation is
`x^(2)-kx+9=0`
Comparing with `ax^(2)+bx+c=0`, we get
a=1, b=-k and c=9
`D=b^(2)-4acgt=0` for real and distinct roots
Therefore, `D=(-k)^(2)-4xx1xx9gt=0`
`impliesk^(2)-36gt=0`
`impliesklt=-6orkgt=6`