Question

write the roots of the equation x 2 +x-p(p+1) where p is constant

Answer 1

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Answer 2

Answer:

The roots of the given equation are $x=p,-p-1$

Solution:

Given:

$\begin{array} { c } { \mathrm { x } ^ { 2 } + \mathrm { x } - \mathrm { p } ( \mathrm { p } + 1 ) = 0 } \\\\ { \mathrm { x } ^ { 2 } + \mathrm { x } - \mathrm { p } ^ { 2 } - \mathrm { p } = 0 } \\\\ { \mathrm { x } ^ { 2 } - \mathrm { p } ^ { 2 } + \mathrm { x } - \mathrm { p } = 0 } \end{array}$

The formula for $a^2-b^2=(a+b)(a-b)$

$\begin{array} { c } { ( \mathrm { x } + \mathrm { p } ) ( \mathrm { x } - \mathrm { p } ) + ( \mathrm { x } - \mathrm { p } ) = 0 } \\\\ { ( \mathrm { x } - \mathrm { p } ) ( \mathrm { x } + \mathrm { p } + 1 ) = 0 } \\\\ { \mathrm { x } - \mathrm { p } = 0 } \\\\ { x = p } \end{array}$

$\begin{array} { l } { x + p + 1 = 0 } \\\\ { x = - p - 1 } \end{array}$

Hence, the roots of the given equation are $x=p, -p-1$