Question

evaluate x^2+99x+127​

Answer 1

$evaluate \: {x}^{2} + 99x \: + 127$

$solution \: > > >$

$let \: given \: quadratic \: polynomial \: be \:$

$p(x) = {x}^{2} + 99x + 127$

$on \: comparing \: p(x) \: with \: a {x}^{2} + by + c \: we \: get \: a = 1 \: \: b = 99 \: and \: c = 127$

$we \: know \: that \: \: \: \:x = \frac{ - b \binom{ + }{ - } \sqrt{ {b }^{2} - 4ac} }{2a} ({ \: by \: quadratic \: formula})$

$= \frac{ - 99 \binom{ + }{ - } \sqrt{ {99}^{2} - 4 \times 1 \times 127 } }{2 \times 1}$

$= \frac{ - 99 \binom{ + }{ - } \sqrt{9801 - 508} }{2}$

$= \frac{ - 99 \binom{ + }{ - } \sqrt{9293} }{2}$

$= \frac{ - 99 \binom{ + }{ - } 964}{2}$

$= \frac{ - 99 + 96.4}{2 \: } \: and \: \frac{ - 99 - 96.4}{2}$

$= \frac{2.6}{2} \: and \: \frac{ - 195.4}{2}$

$= 1.3 \: and \: - 97.7$

$hence \: both \: zeroes \: of \: the \: given \: quadratic \: polynomial \: p(x) \: are \: negative \:$

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