Math, asked by bhaiarota07, 10 months ago

If a and ß are zeroes of the polynomial 2x² + 3x – 6, then find the value of a +B2 - aß.​

Answers

Answered by irshadsyed281
2

\bold{\blue{\underline{\red{G}\pink{iv}\green{en}\purple{:-}}}}\\\\

  • P(x) = 2x² + 3x – 6
  • α and β are the zeros

\bold{\blue{\underline{\red{Zeros}\pink{\:of\:}\green{\:the\:Quadratic\:equation}\purple{:-}}}}

  • Zeros of the quadratic equation is the value of the variable for which when replaced in the quadratic equations the value for the quadratic equation changes to zero.

\bold{\blue{\underline{\red{General}\pink{\:form \:of }\green{\:Quadratic\:equation}\purple{:-}}}}

  • ax² + bx + c = 0

\bold{\blue{\underline{\red{Q}\pink{uest}\green{ion}\purple{:-}}}}\\

  • To find the value of α + β² - αβ

\bold{\blue{\underline{\red{S}\pink{olut}\green{ion}\purple{:-}}}}

  • 2x² + 3x 6
  • a = 2
  • b = 3
  • c = -6
  • Quadratic formula = \bold{\frac{-b\: \pm\:{\sqrt{b^{2}\:-\:4ac} }}{2a}}  

  • α = \bold{\frac{-3\: \pm\:{\sqrt{3^{2}\:-\:4(2)(-6)} }}{2(3)}}  

  • α = \bold{\frac{-3\: \pm\:{\sqrt{9\:+\:48} }}{6}}  

  • α = \bold{\frac{-3\: +\:{\sqrt{57} }}{6}}

  • β = \bold{\frac{-3\: -\\\:{\sqrt{57} }}{6}}

   To find the value of α + β² - αβ :

  • \bold{\frac{-3\: +\:{\sqrt{57} }}{6}}  +  \bold({\frac{-3\: -\\\:{\sqrt{57} }}{6})}^2   -    \bold{\frac{-3\: +\\\:{\sqrt{57} }}{6}} \: \times\: \bold{\frac{-3\: -\\\:{\sqrt{57} }}{6}}

  • \bold{\frac{-3\: +\:{\sqrt{57} }}{6}}  + \bold{\frac{9\: +\:57 }{36}} - \bold{\frac{9\: -\:57 }{36}}

  • \bold{\frac{-3\: -\:\sqrt{57} }{6}}  +  \bold{\frac{18}{36} }

  • \bold{\frac{-3\: -\:\sqrt{57} }{6}} + \bold{\frac{3}{6} }

  • \bold{\frac{-3\: -\:\sqrt{57}\:+\:3 }{6}}

  • \bold{\frac{ -\:\sqrt{57} }{6}}
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