1.

Find the zeroes of the qudratic polynomia :6x²-x-2​

Answer»

Step-by-step explanation:

  • 6x² - X - 2 = 0
  • 6x² + 4X - 3x - 2 = 0
  • 2x ( 3x + 2 ) - 1 ( 3x - 2 ) = 0
  • ( 2x - 1 ) + ( 3x + 2 ) = 0

Two ZEROES :-

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→ 2x - 1 = 0

→ 2x = 1

→ x = ½

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→ 3x + 2 = 0

→ 3x = -2

→ x = \frac{-2}{3}

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\alpha = ½ and \beta = \frac{-2}{3}

We know that

\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha }

\frac{ \frac{1}{2} }{ \frac{ - 2}{3} } + \frac{ \frac{ - 2}{3} }{ \frac{1}{2} }

= > ( \frac{1}{2} \times \frac{3}{ - 2} ) \: + \: ( \: \frac{ - 2}{3} \times \frac{2}{1} )

= > \frac{ - 3}{4} \: + \: \frac{4}{3}

= > \frac{ - 9 - 16}{12} \\ \\ = > \frac{ - 25}{12}

Thus ,

\alpha + \beta = \frac{-25}{12}


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