Respuestas
Respuesta dada por:
2
Desarrollo:
y ![r_{2} = \frac{2}{5} r_{2} = \frac{2}{5}](https://tex.z-dn.net/?f=+r_%7B2%7D+%3D+%5Cfrac%7B2%7D%7B5%7D+)
Sabemos que:
![a x^{2} +bx+c a x^{2} +bx+c](https://tex.z-dn.net/?f=a+x%5E%7B2%7D+%2Bbx%2Bc)
también:
![r_{1}+ r_{2} =- \frac{b}{a} r_{1}+ r_{2} =- \frac{b}{a}](https://tex.z-dn.net/?f=+r_%7B1%7D%2B+r_%7B2%7D+%3D-++%5Cfrac%7Bb%7D%7Ba%7D+)
![( r_{1} )( r_{2} )= \frac{c}{a} ( r_{1} )( r_{2} )= \frac{c}{a}](https://tex.z-dn.net/?f=%28+r_%7B1%7D+%29%28+r_%7B2%7D+%29%3D+%5Cfrac%7Bc%7D%7Ba%7D+)
Donde
y
son raíces
Por tanto:
![r_{1} + r_{2} =- \frac{b}{a} = \frac{1}{3} + \frac{2}{5} = \frac{11}{15} r_{1} + r_{2} =- \frac{b}{a} = \frac{1}{3} + \frac{2}{5} = \frac{11}{15}](https://tex.z-dn.net/?f=+r_%7B1%7D+%2B+r_%7B2%7D+%3D-+%5Cfrac%7Bb%7D%7Ba%7D+%3D+%5Cfrac%7B1%7D%7B3%7D+%2B+%5Cfrac%7B2%7D%7B5%7D+%3D+%5Cfrac%7B11%7D%7B15%7D+)
![- \frac{b}{a} = \frac{11}{15} - \frac{b}{a} = \frac{11}{15}](https://tex.z-dn.net/?f=-+%5Cfrac%7Bb%7D%7Ba%7D+%3D+%5Cfrac%7B11%7D%7B15%7D+)
![b=- \frac{11a}{15} b=- \frac{11a}{15}](https://tex.z-dn.net/?f=b%3D-+%5Cfrac%7B11a%7D%7B15%7D+)
![( r_{1} )( r_{2} )= \frac{c}{a} = ( \frac{1}{3} )( \frac{2}{5}) = \frac{2}{15} ( r_{1} )( r_{2} )= \frac{c}{a} = ( \frac{1}{3} )( \frac{2}{5}) = \frac{2}{15}](https://tex.z-dn.net/?f=%28+r_%7B1%7D+%29%28+r_%7B2%7D+%29%3D+%5Cfrac%7Bc%7D%7Ba%7D+%3D+%28+%5Cfrac%7B1%7D%7B3%7D+%29%28+%5Cfrac%7B2%7D%7B5%7D%29+%3D+%5Cfrac%7B2%7D%7B15%7D+)
![\frac{c}{a} = \frac{2}{15} \frac{c}{a} = \frac{2}{15}](https://tex.z-dn.net/?f=+%5Cfrac%7Bc%7D%7Ba%7D+%3D+%5Cfrac%7B2%7D%7B15%7D+)
![c= \frac{2a}{15} c= \frac{2a}{15}](https://tex.z-dn.net/?f=c%3D+%5Cfrac%7B2a%7D%7B15%7D+)
Remplazando:
![a x^{2} +bx+c a x^{2} +bx+c](https://tex.z-dn.net/?f=a+x%5E%7B2%7D+%2Bbx%2Bc)
![a x^{2} - \frac{11a}{15} x+ \frac{2a}{15}=0 a x^{2} - \frac{11a}{15} x+ \frac{2a}{15}=0](https://tex.z-dn.net/?f=a+x%5E%7B2%7D+-+%5Cfrac%7B11a%7D%7B15%7D+x%2B+%5Cfrac%7B2a%7D%7B15%7D%3D0+)
Reduciendo:
![15 x^{2} -11x+2=0 15 x^{2} -11x+2=0](https://tex.z-dn.net/?f=15+x%5E%7B2%7D+-11x%2B2%3D0)
Sabemos que:
también:
Donde
Por tanto:
Remplazando:
Reduciendo:
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