\( x_{{1,2}} = \frac{{ – b\ \pm\ \sqrt {b^{2}\ –\ 4ac} }}{{2a}} = \frac{{ – 2p\ \pm\ \sqrt 4 }}{{2\left( {p^{2}\ –\ 1} \right)}} = \frac{{ – 2p\ \pm\ 2}}{{2\left( {p^{2}\ –\ 1} \right)}} = \frac{{2\left( { – p\ \pm\ 1} \right)}}{{2\left( {p^{2}\ –\ 1} \right)}} = \frac{{ – p\ \pm\ 1}}{{p^{2}\ –\ 1}} \)
\( x_{1} = \frac{{ – p\ +\ 1}}{{\left( {p\ –\ 1} \right)\left( {p\ +\ 1} \right)}} = \frac{{ – 1\left( {p\ –\ 1} \right)}}{{\left( {p\ –\ 1} \right)\left( {p\ +\ 1} \right)}} = – \frac{1}{{p\ +\ 1}} \)
\( x_{2} = \frac{{ – p\ –\ 1}}{{p^{2}\ –\ 1}} = \frac{{ – 1\left( {p\ +\ 1} \right)}}{{\left( {p\ –\ 1} \right)\left( {p\ +\ 1} \right)}} = \frac{{ – 1}}{{p\ –\ 1}} = \frac{1}{{ – 1\left( {p\ –\ 1} \right)}} = \frac{1}{{1\ –\ p}} \)
\( \large K_{3} = \left \{ { - \Large \frac{1}{{p+1}}\large ;\Large \frac{1}{{1-p}}\large } \right \} \)
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