Question

Una solución al p al cuadrado+11p+10=0 es

Answer 1

SOLUCIÓN

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                         $Sea\:la\:ecuaci\'on\:cuadr\'atica:\\\\\\ax^2+bx+c=0\\\\\\Por\:f\'ormula\:general\:tenemos\:que:\\\\\\\mathrm{\boxed{x_{1,2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}}\\\\\\Del\:problema:\\\\\\a=1,b=11,c=10\\\\\\ Entonces\:reemplazamos\\\\\\p_{1,2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\p_{1,2}=\dfrac{-(11)\pm \sqrt{(11)^2-[4(1)(10)]}}{2(1)}\\\\\\p_{1,2}=\dfrac{-11\pm \sqrt{121-(40)}}{2}\\\\\\p_{1,2}=\dfrac{-11\pm \sqrt{81}}{2}\\\\\\p_{1,2}=\dfrac{-11\pm9}{2}$

                      $\Rightarrow\:p_{1}=\dfrac{-11+9}{2}\\\\\\p_{1}=\dfrac{-2}{2}\\\\\\\boxed{\boldsymbol{p_{1}=-1}}\\\\\\\Rightarrow\:p_{2}=\dfrac{-11-9}{2}\\\\\\p_{2}=\dfrac{-20}{2}\\\\\\\boxed{\boldsymbol{p_{2}=-10}}$

Answer 2

Respuesta:

$p = \frac{ - 11 + - \sqrt{11 {}^{2} } - 4 \times 1 \times 10 }{2 \times 1}$

$p = \frac{ - 11 + - \sqrt{121 - 40} }{2}$

$p = \frac{ - 11 + - \sqrt{81} }{2}$

$p = \frac{ - 11 + - 9}{2}$

$p1 = \frac{ - 11 + 9}{2} = \frac{ - 2}{2} = - 1$

$p2 = \frac{ - 11 - 9}{2} = \frac{ - 20}{2} = - 10$