Determina las raíces de las siguientes ecuaciones cuadráticas 1) x2 – 3x = 0 2) 6x2 + 42x = 0 3) x2.... Question from @camiladiaz110301

November 2018 1 181 Report

Determina las raíces de las siguientes ecuaciones cuadráticas
1) x2 – 3x = 0
2) 6x2 + 42x = 0
3) x2 + ax = 0
4) x2 – 18x + 80 = 0
5) x2 – 4x – 96 = 0
6) x2 – 17x + 52 = 0
7) x2 – 7x – 120 = 0
8) 4x2 + 5x – 6 = 0
9) 6x2 + 5x – 1 = 0
10) 3x2 – 10x – 25 = 0
11) 7x2 – 16x + 9 = 0
12) x2 – 5ax + 6a2 = 0
me ayudan pliss

Piscis04 Para determinar las raíces de las funciones cuadráticas se resuelve así! :

$Forma \ Polin\'omica\ \to ax^2+bx+c=0 \\ \\
1) Ecuaci\'on \ incompleta \quad\to x^2-3x= 0 \\ \\ Se \ aplica \ factor \
com\'un \\ \\ x^2-3x= 0 \\ \\ x(x-3) =0 \\ \\ x=0\qquad\qquad x= 3 \\ \\ Las \
ra\'ices\ son \ \boxed{ x_1=0\ y\ x_2= 3 }$
------------------------------------
$2) 6x^2+42x= 0 \\ \\ Se \ aplica \ factor \
com\'un \\ \\ x^2-3x= 0 \\ \\ 6x(x+7) =0 \\ \\ 6x=0\to x= 0\qquad\qquad x+7= 0
\to x= -7\\ \\ Las \ ra\'ices\ son \ \boxed{ x_1=0\ y\ x_2= -7 }$
-----------------------------------
$3) x^2+ax= 0 \\ \\ Se \ aplica \
factor \ com\'un \\ \\ x^2+ax= 0 \\ \\ x(x+a) =0 \\ \\ x=0\qquad\qquad x+a=0\to
x=a \\ \\ Las \ ra\'ices\ son \ \boxed{ x_1=0\ y\ x_2= a }$
----------------------------------
$4) x^2-18x+80= 0 \\ \\ Se \ aplica \ Bascara\to
x_{1\ y\ 2}= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} , entonces \\ \\ x^2-18x+80= 0
\\ \\ x_{1\ y\ 2}= \dfrac{-(-18)\pm \sqrt{(-18)^2-4(1)(80)} }{2(1)}\qquad a=
1\quad b=-18\quad c=80 \\ \\ x_{1\ y\ 2}= \dfrac{+18\pm \sqrt{324-320} }{2} \\
\\ x_{1\ y\ 2}= \dfrac{+18\pm \sqrt{4} }{2} \\ \\x_{1\ y\ 2}=\dfrac{18\pm2 }{2}
\\ \\ \\x_{1}=\dfrac{18+2 }{2} \qquad x_{1}=\dfrac{20 }{2} \qquad \boxed{x_1=
10}$
$\\ \\ \\x_{2}=\dfrac{18-2 }{2} \qquad
x_{2}=\dfrac{16}{2} \qquad \boxed{x_2= 4}$
-----------------------------------
$5) x^2-4x-96= 0 \\ \\ Se \ aplica \
Bascara\to x_{1\ y\ 2}= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} , entonces \\ \\ \\
\\ x_{1\ y\ 2}= \dfrac{-(-4)\pm \sqrt{(-4)^2-4(1)(-96)} }{2(1)}\qquad a= 1\quad
b=-4\quad c=-96 \\ \\ x_{1\ y\ 2}= \dfrac{+4\pm \sqrt{16+384} }{2} \\ \\ x_{1\
y\ 2}= \dfrac{+4\pm \sqrt{400} }{2} \\ \\x_{1\ y\ 2}=\dfrac{4\pm20 }{2} \\ \\
\\x_{1}=\dfrac{4+20 }{2} \qquad x_{1}=\dfrac{24 }{2} \qquad \boxed{x_1=6}\\ \\
\\x_{2}=\dfrac{4-20 }{2} \qquad x_{2}=\dfrac{-16 }{2} \qquad
\boxed{x_2=-8}$
-----------------------------------
$6) x^2-17x+52= 0 \\ \\ Se \ aplica \
Bascara\to x_{1\ y\ 2}= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} , entonces \\ \\ \\
\\ x_{1\ y\ 2}= \dfrac{-(-17)\pm \sqrt{(-17)^2-4(1)(52)} }{2(1)}\qquad a=
1\quad b=-17\quad c=52 \\ \\ x_{1\ y\ 2}= \dfrac{+17\pm \sqrt{289-208} }{2} \\
\\ x_{1\ y\ 2}= \dfrac{17\pm \sqrt{81} }{2} \\ \\x_{1\ y\ 2}=\dfrac{17\pm9 }{2}
\\ \\ \\x_{1}=\dfrac{17+9 }{2} \qquad x_{1}=\dfrac{26 }{2} \qquad
\boxed{x_1=13}\\ \\ \\x_{2}=\dfrac{17-9 }{2} \qquad x_{2}=\dfrac{8 }{2} \qquad
\boxed{x_2=4}$
-------------------------------
$7) x^2-7x-120= 0 \\ \\ Se \ aplica \
Bascara\to x_{1\ y\ 2}= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} , entonces \\ \\ \\
\\ x_{1\ y\ 2}= \dfrac{-(-7)\pm \sqrt{(-7)^2-4(1)(-120)} }{2(1)}\qquad a=
1\quad b=-7\quad c=-120 \\ \\ x_{1\ y\ 2}= \dfrac{+7\pm \sqrt{49+480} }{2} \\
\\ x_{1\ y\ 2}= \dfrac{7\pm \sqrt{529} }{2} \\ \\x_{1\ y\ 2}=\dfrac{7\pm23 }{2}
\\ \\ \\x_{1}=\dfrac{7+23 }{2} \qquad x_{1}=\dfrac{30 }{2} \qquad
\boxed{x_1=15}\\ \\ \\x_{2}=\dfrac{7-23 }{2} \qquad x_{2}=\dfrac{-16 }{2}
\qquad \boxed{x_2=-8}$
---------------------------
$8) 4x^2+5x-6= 0 \\ \\ Se \ aplica \
Bascara\to x_{1\ y\ 2}= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} , entonces \\ \\ \\
\\ x_{1\ y\ 2}= \dfrac{-(5)\pm \sqrt{(5)^2-4(4)(-6)} }{2(4)}\qquad a= 4\quad
b=5\quad c=-6 \\ \\ x_{1\ y\ 2}= \dfrac{-5\pm \sqrt{25+96} }{8} \\ \\ x_{1\ y\
2}= \dfrac{-5\pm \sqrt{121} }{8} \\ \\x_{1\ y\ 2}=\dfrac{-5\pm11 }{8} \\ \\
\\x_{1}=\dfrac{-5+11 }{8} \qquad x_{1}=\dfrac{6}{8} \qquad \boxed{x_1=
\frac{3}{4} }\\ \\ \\x_{2}=\dfrac{-5-11}{8} \qquad x_{2}=\dfrac{-16 }{8} \qquad
\boxed{x_2=-2}$

----------------------------------
$9) 6x^2+5x-1= 0 \\ \\ Se \ aplica \
Bascara\to x_{1\ y\ 2}= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} , entonces \\ \\ \\
\\ x_{1\ y\ 2}= \dfrac{-(5)\pm \sqrt{(5)^2-4(6)(-1)} }{2(6)}\qquad a=6\quad
b=5\quad c=-1 \\ \\ x_{1\ y\ 2}= \dfrac{-5\pm \sqrt{25+24} }{12} \\ \\ x_{1\ y\
2}= \dfrac{-5\pm \sqrt{49} }{12} \\ \\x_{1\ y\ 2}=\dfrac{-5\pm7 }{12} \\ \\
\\x_{1}=\dfrac{-5+7 }{12} \qquad x_{1}=\dfrac{2 }{12} \qquad
\boxed{x_1=\frac{1}{6}}\\ \\ \\x_{2}=\dfrac{-5-7}{12} \qquad x_{2}=\dfrac{-12}{12}
\qquad \boxed{x_2=-1}$
---------------------------------
$10)3x^2-10x-25= 0 \\ \\ Se \ aplica \
Bascara\to x_{1\ y\ 2}= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} , entonces \\ \\ \\
x_{1\ y\ 2}= \dfrac{-(-10)\pm \sqrt{(-10)^2-4(3)(-25)} }{2(3}\quad a=3\quad
b=-10\quad c=-25 \\ \\ x_{1\ y\ 2}= \dfrac{10\pm \sqrt{100+300} }{6} \\ \\
x_{1\ y\ 2}= \dfrac{10\pm \sqrt{400} }{6} \\ \\x_{1\ y\ 2}=\dfrac{10\pm20 }{6}
\\ \\ x_{1}=\dfrac{10+20 }{6} \qquad x_{1}=\dfrac{30}{6} \qquad \boxed{x_1=5}\\
\\ x_{2}=\dfrac{10-20}{6} \quad x_{2}=\dfrac{-10}{6} \quad
\boxed{x_2=-\frac{-5}{3}}$
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Te agrego el 11) y el 12) en adjuntos porque el lugar de respuesta no alcanzo para hacerlo acá