Question

find the roots of equation x^2+x-p(p+1),where p is constant​

Answer 1

Answer:

$x = \frac{ - 1 + (2p + 1)}{2}$

$x = \frac{ - 1 - (2p + 1}{2}$

Step-by-step explanation:

${x}^{2} + x - p(p + 1)$

Using quadratic formula:

$x = \frac{ - 1 (+ -) \sqrt{1 + 4( {p}^{2} + p } }{2}$

$x = \frac{ - 1( + - ) \sqrt{4 {p}^{2} + 4p + 1 } }{2}$

$x = \frac{ - 1( + - ) \sqrt{ {2p + 1}^{2} } }{2}$

$x = \frac{ - 1( + - )(2p + 1)}{2}$

(+-) means plus or minus.

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