Question 1132103


The roots of the equation {{{x^2 + 2px + q = 0}}} differ by{{{ 2}}}. Show that
{{{p^2 = 1 + q }}}

{{{x^2 + 2px + q = 0}}}

Comparing with {{{ax^2 + bx + c = 0}}} we have {{{a = 1}}}, {{{b = 2p}}}, {{{c = q}}}

Let {{{alpha}}} and {{{beta}}}  be the roots of given quadratic equation.

{{{alpha -beta  = 2}}} ......(i) [Given]

 {{{alpha+ beta = -b/a = -2p/1 = -2p}}}

Also, {{{alpha*beta= c/a = q/1 = q}}}

We know that,

{{{(alpha -beta )^2 = (alpha+beta)^2  -4alpha*beta}}}..substitute values above

{{{ (2)^2 = (-2p)^2 -4 (q)}}}

{{{4 = 4p^2 -4q}}}...simplify

{{{1 = p^2 -q}}}

{{{ p^2 = 1+ q}}}

Hence proved.