Matemáticas, pregunta formulada por EmmaValentina868, hace 5 meses

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C. 2x2 + 9x + 10 = 0
D. 5x2 + 7x + 2 = 0
el x2 significa al cuadrado
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Respuestas a la pregunta

Contestado por roycroos

╔═════════════════════════════════════════════╗ $\center \mathrm{Una ecuaci\'on cuadr\'atica de la forma:}$

$\mathrm{ax^2 + bx + c=0\:\:donde\:\: a \neq 0}$

$\center \mathrm{Poseer\'a 2 soluciones x_1} \:\mathrm{y\:x_2,\:las\:cuales\:determinaremos\:por:}$

$\boldsymbol{{\mathrm{x_{1,2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}}} \Rightarrow \boxed{\mathrm{F\'ormula\:general}}$ ╚═════════════════════════════════════════════╝

C.

En el problema

$\mathsf{\underbrace{\boldsymbol{2}}_{a}x^2\:+\:\underbrace{\boldsymbol{9}}_{b}x\:+\:\underbrace{\boldsymbol{10}}_{c}=0}$

Entonces a = 2, b = 9, c = 10

Reemplazamos los coeficientes en la fórmula general:

$\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-(9) \pm \sqrt{(9)^2 - [4(2)(10)]}}{2(2)}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-9 \pm \sqrt{81 - (80)}}{4}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-9 \pm \sqrt{1}}{4}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-9 \pm 1}{4}}$

$\center \Rightarrow\:\mathrm{x_{1}} \mathrm{= \dfrac{-9 + 1}{4}}\\\\\\\center \mathrm{x_{1}} \mathrm{= \dfrac{-8}{4}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{1}} \mathrm{= -2}}}}$ $\center \Rightarrow\:\mathrm{x_{2}} \mathrm{= \dfrac{-9 - 1}{4}}\\\\\\\center \mathrm{x_{2}} \mathrm{= \dfrac{-10}{4}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{2}} \mathrm{= -2.5}}}}$

D.

En el problema

$\mathsf{\underbrace{\boldsymbol{5}}_{a}x^2\:+\:\underbrace{\boldsymbol{7}}_{b}x\:+\:\underbrace{\boldsymbol{2}}_{c}=0}$

Entonces a = 5, b = 7, c = 2

Reemplazamos los coeficientes en la fórmula general:

$\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-(7) \pm \sqrt{(7)^2 - [4(5)(2)]}}{2(5)}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-7 \pm \sqrt{49 - (40)}}{10}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-7 \pm \sqrt{9}}{10}}\\\\\\\center \mathrm{x_{1,2}} \mathrm{= \dfrac{-7 \pm 3}{10}}$

$\center \Rightarrow\:\mathrm{x_{1}} \mathrm{= \dfrac{-7 + 3}{10}}\\\\\\\center \mathrm{x_{1}} \mathrm{= \dfrac{-4}{10}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{1}} \mathrm{= -0.4}}}}$ $\center \Rightarrow\:\mathrm{x_{2}} \mathrm{= \dfrac{-7 - 3}{10}}\\\\\\\center \mathrm{x_{2}} \mathrm{= \dfrac{-10}{10}}\\\\\\\center \boxed{\boxed{\boldsymbol{\mathrm{x_{2}} \mathrm{= -1}}}}$