Question

Show that the equation x square + px - 1 = 0 has
real and distinct roots for all real values of p.​

Answer 1

Answer:

${x}^{2} + px - 1 = 0$

Find D

$d = { {p}^{2} + 4 }$

Now

${ {p}^{2} + 4 } > 0 \: \: \: \: \:p \in \: r$

So it has real and distinct roots