Estas ecuaciones son las que tienen la forma \(ax^{2}+bx+c=0\) siendo \(a\neq 0;b\neq 0;c\neq 0\).
La formula es la siguiente \(\boxed{x=\dfrac {-b\pm \sqrt {b^{2}-4ac}} {2a}}\)
Podemos tener dos soluciones de x, una sola solución o ninguna solución real.
Aquí os dejo varios ejemplos.
\(\bullet x^{2}-10x+21=0\)
\(x=\dfrac {10\pm \sqrt {10^{2}-4\cdot1\cdot21}} {2\cdot 1}=\dfrac {10\pm \sqrt {100-84}} {2}\)
\(\boxed{x_{1}=7\\x_{2}=3}\)
\(\bullet x^{2}-10x+25=0\)
\(x=\dfrac {10\pm \sqrt {10^{2}-4\cdot1\cdot25}} {2\cdot 1}=\dfrac {10\pm \sqrt {100-100}} {2}\)
\(\boxed{x_{1}=5\\x_{2}=5}\)
\(\bullet x^{2}+9x+40=0\)
\(x=\dfrac {-9\pm \sqrt {9^{2}-4\cdot1\cdot40}} {2\cdot 1}=\dfrac {-9\pm \sqrt {81-160}} {2}\)
\(\boxed{Sin\\solución\\real }\)
\(\bullet x^{2}-5x+6=0\)
\(x=\dfrac {5\pm \sqrt {5^{2}-4\cdot1\cdot6}} {2\cdot 1}=\dfrac {5\pm \sqrt {25-24}} {2}\)
\(\boxed{x_{1}=3\\x_{2}=2}\)
\(\bullet -5x^{2}+13x+6=0\)
\(5x^{2}-13x-6=0\)
\(x=\dfrac {13\pm \sqrt {13^{2}-4\cdot5\cdot6}} {2\cdot 5}=\dfrac {13\pm \sqrt {169--120}} {10}\)
\(\boxed{x_{1}=3\\x_{2}=-\dfrac {2} {5}}\)
miércoles, 28 de mayo de 2014
Ecuaciones de primer grado con denominadores. Incógnitas en el denominador
Para realizar estas ecuaciones multiplicaremos por el m.c.m. ambas partes de la igualdad. Aquí os dejo unos ejemplos resueltos de este tipo de ejercicios.
\(\bullet \dfrac {1} {x}+\dfrac {1} {2}=3\)
\(2x \left( \dfrac {1} {x}+\dfrac {1} {2}\right) =3\cdot 2x\)
\(2+x=6x\)
\(\boxed{x=\dfrac{2} {5}}\)
\(\bullet \dfrac {1} {2}-\dfrac {1} {x}=\dfrac {2} {5}\)
\(10x \left( \dfrac {1} {2}-\dfrac {1} {x}\right) =10x\cdot \dfrac {2} {5}\)
\(5x-10=4x\)
\(\boxed{x=10}\)
\(\bullet \dfrac {2} {x-1}-1=\dfrac {4x} {x-1}\)
\(\left( x-1\right) \left( \dfrac {2} {x-1}-1\right) =\left( x-1\right) \dfrac {4x} {x-1}\)
\(2-x+1=4x\)
\(5x=3\)
\(\boxed{x=\dfrac{3} {5}}\)
\(\bullet \dfrac {1} {x-4}+3=\dfrac {1} {5}\)
\(5\left( x-4\right) \left( \dfrac {1} {x-4}+3\right) =5\left( x-4\right) \cdot 1\)
\(5+15x-60=5x-20\)
\(10x=35\)
\(\boxed{x=\dfrac{7} {2}}\)
\(\bullet \dfrac {4x} {x-1}+\dfrac {2} {5\left( x-1\right) }=\dfrac {3} {5}\)
\(5\left( x-1\right) \left( \dfrac {4x} {x-1}+\dfrac {2} {5\left( x-1\right) }\right) =5\left( x-1\right) \cdot \dfrac {3} {5}\)
\(20x+2=3x-3\)
\(17x=-5\)
\(\boxed{x=-\dfrac{5} {17}}\)
martes, 27 de mayo de 2014
Ecuaciones de primer grado con denominadores. Incógnitas sólo en el numerador
Como en las otras ecuaciones de primer grado Juntaremos los monomios con x a un lado de la igualdad y los sin x a otro lado
\(\bullet \dfrac {1} {5}\left( 2+5x\right) =\dfrac {1} {2}\left( x-\dfrac {1} {5}\right) \)
\(\dfrac {2} {5}+x=\dfrac {1} {2}x-\dfrac {1} {10}\)
\(4+10x=5x-1\)
\(5x=-5\)
\(\boxed{x=-1}\)
\(\bullet 5\left( \dfrac {x} {4}-\dfrac {1} {10}\right) =\dfrac {1} {2}\left( 3x-\dfrac {1} {2}\right) \)
\(\dfrac {5} {4}x-\dfrac {1} {2}=\dfrac {3} {2}x-\dfrac {1} {4}\)
\(5x-2=6x-1\)
\(\boxed{x=-1}\)
\(\bullet \dfrac {3x} {2}-1=\dfrac {3x+2} {4}\)
\(6x-4=3x+2\)
\(3x=6\)
\(\boxed{x=2}\)
\(\bullet \dfrac {x} {5}-\dfrac {x+2} {15}=\dfrac {x} {3}\)
\(3x-x-2=5x\)
\(\boxed{x=-\dfrac {2} {3}}\)
\(\bullet \dfrac {1-x} {3}-\dfrac {x-1} {12}=\dfrac {3x-1} {4}\)
\(4-4x-x+1=9x-3\)
\(-14x=-8\)
\(\boxed{x=-\dfrac {4} {7}}\)
Dejadme un comentario o algún tipo de problema que no sepas resolver.
\(\bullet \dfrac {1} {5}\left( 2+5x\right) =\dfrac {1} {2}\left( x-\dfrac {1} {5}\right) \)
\(\dfrac {2} {5}+x=\dfrac {1} {2}x-\dfrac {1} {10}\)
\(4+10x=5x-1\)
\(5x=-5\)
\(\boxed{x=-1}\)
\(\bullet 5\left( \dfrac {x} {4}-\dfrac {1} {10}\right) =\dfrac {1} {2}\left( 3x-\dfrac {1} {2}\right) \)
\(\dfrac {5} {4}x-\dfrac {1} {2}=\dfrac {3} {2}x-\dfrac {1} {4}\)
\(5x-2=6x-1\)
\(\boxed{x=-1}\)
\(\bullet \dfrac {3x} {2}-1=\dfrac {3x+2} {4}\)
\(6x-4=3x+2\)
\(3x=6\)
\(\boxed{x=2}\)
\(\bullet \dfrac {x} {5}-\dfrac {x+2} {15}=\dfrac {x} {3}\)
\(3x-x-2=5x\)
\(\boxed{x=-\dfrac {2} {3}}\)
\(\bullet \dfrac {1-x} {3}-\dfrac {x-1} {12}=\dfrac {3x-1} {4}\)
\(4-4x-x+1=9x-3\)
\(-14x=-8\)
\(\boxed{x=-\dfrac {4} {7}}\)
Dejadme un comentario o algún tipo de problema que no sepas resolver.
Descomposición en factores y simplificación
Las siguientes expresiones podrán descomponerse mediante productos notables o sacando factor común
\(\bullet \dfrac {x^{2}-9} {x^{2}-6x+9}=\dfrac {\left( x+3\right)\enclose{updiagonalstrike} {\left( x-3\right)} } {\left( x-3\right) ^\enclose{updiagonalstrike}{{2}}}=\boxed{\dfrac {x+3} {x-3}}\)
\(\bullet \dfrac {3x^{2}+3} {3x^{2}-3}=\dfrac {\enclose{updiagonalstrike}{3}\enclose{updiagonalstrike}{\left( x+1\right) }} {\enclose{updiagonalstrike}{3}\enclose{updiagonalstrike}{\left( x+1\right)} \left( x-1\right) }=\boxed{\dfrac {1} {x-1}}\)
\(\bullet \dfrac {x^{2}+2x+1} {5x^{2}+5x}=\dfrac {\left( x+1\right) ^\enclose{updiagonalstrike}{{2}}} {5\enclose{updiagonalstrike}{\left( x+1\right)} }=\boxed{\dfrac {x+1} {5x}}\)
\(\bullet \dfrac {5x+15} {x^{2}+6x+9}=\dfrac {5\enclose{updiagonalstrike}{\left( x+3\right)} } {\left( x+3\right) ^\enclose{updiagonalstrike}{{2}}}=\boxed{\dfrac {5} {x+3}}\)
\(\bullet \dfrac {3x^{2}+6x+3} {5x^{2}+5x}=\dfrac {3\left( x+1\right) ^\enclose{updiagonalstrike}{{2}}} {5x\enclose{updiagonalstrike}{\left( x+1\right)} }=\boxed{\dfrac {3\left( x+1\right) } {5x}}\)
Si tenéis alguna duda, dejadme un mensaje muchas gracias
\(\bullet \dfrac {x^{2}-9} {x^{2}-6x+9}=\dfrac {\left( x+3\right)\enclose{updiagonalstrike} {\left( x-3\right)} } {\left( x-3\right) ^\enclose{updiagonalstrike}{{2}}}=\boxed{\dfrac {x+3} {x-3}}\)
\(\bullet \dfrac {3x^{2}+3} {3x^{2}-3}=\dfrac {\enclose{updiagonalstrike}{3}\enclose{updiagonalstrike}{\left( x+1\right) }} {\enclose{updiagonalstrike}{3}\enclose{updiagonalstrike}{\left( x+1\right)} \left( x-1\right) }=\boxed{\dfrac {1} {x-1}}\)
\(\bullet \dfrac {x^{2}+2x+1} {5x^{2}+5x}=\dfrac {\left( x+1\right) ^\enclose{updiagonalstrike}{{2}}} {5\enclose{updiagonalstrike}{\left( x+1\right)} }=\boxed{\dfrac {x+1} {5x}}\)
\(\bullet \dfrac {5x+15} {x^{2}+6x+9}=\dfrac {5\enclose{updiagonalstrike}{\left( x+3\right)} } {\left( x+3\right) ^\enclose{updiagonalstrike}{{2}}}=\boxed{\dfrac {5} {x+3}}\)
\(\bullet \dfrac {3x^{2}+6x+3} {5x^{2}+5x}=\dfrac {3\left( x+1\right) ^\enclose{updiagonalstrike}{{2}}} {5x\enclose{updiagonalstrike}{\left( x+1\right)} }=\boxed{\dfrac {3\left( x+1\right) } {5x}}\)
Si tenéis alguna duda, dejadme un mensaje muchas gracias
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