Question

Show that the discriminant of (k+1)x²+4kx+2=0 defines a quadratic function of k.

Answer 1

Explanation:

Given

`\left ( k+1\right )x^2+4kx+2=0`

For Discriminant of ax^2+bx+c=0

`D=√(b^2-4ac)`

here

`D=√(\left ( 4k\right )^2-4\left ( k+1\right )\left ( 2\right ))`

`D=√(16k^2-8k-8)>0` for real roots

so
`16k^2-8k-8>0`

`2k^2-k-1>0` is a form of Quadratic equation