Question

2x+3x+1=0 find roots alpha ,beta
alpha +beta /beta +alpha


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Answer 1

Question :

  2x² + 3x + 1 = 0

Find :-

• roots α,β
• α+β/αβ

Answer :

• The roots of the quadratic equation are -1 and -1/2
• α+β/αβ = -3

Given :

• 2x² + 3x + 1 = 0

To find :

• α,β
• α+β/αβ

Solution :

   Given quadratic equation,

        2x² + 3x + 1 = 0

α and β are the roots of the given quadratic equation.

${\text{\underline{\underline{\bf Finding the roots :}}}$

 The given quadratic equation is of the form ax² + bx + c = 0

   a = 2, b = 3, c = 1

To solve it using factorization method,

we must know the sum - product pattern

>> Find the product of quadratic term [ax²] and constant term [c]

 = 2x² × 1

 = 2x²

>> find the factors of "2x²" in pairs

  $=> 2x \times x$

>> Since we got only one

  check if the pair adds to get linear term [bx]

>> So, split 3x as 2x and x

   2x² + 3x + 1 = 0

  2x² + 2x + x + 1 = 0

>> Find the common factor,

  2x(x+1) + 1(x+1) = 0

    (x + 1) (2x + 1) = 0

=>   x+1 = 0 ; x = -1

     2x+1 = 0 ; x = -1/2

     ∴ -1 and -1/2 are the roots of the quadratic equation 2x²+3x+1 = 0

$\underline{\underline{\bf Finding \ \frac{\alpha + \beta}{\alpha \beta} \ : }}$

      $\frac{\alpha + \beta}{\alpha \beta} =\frac{sum \ of \ roots}{product \ of \ roots} \\\\ \frac{\alpha + \beta}{\alpha \beta} = \frac{-1+(\frac{-1}{2})}{(-1)(\frac{-1}{2})} \\\\ \frac{\alpha + \beta}{\alpha \beta} = \frac{-1-\frac{1}{2} }{\frac{1}{2} } \\\\ \frac{\alpha + \beta}{\alpha \beta} = \frac{\frac{-2-1}{2} }{\frac{1}{2} } \\\\ \frac{\alpha + \beta}{\alpha \beta} =\frac{-3}{1} \\\\ \boxed{\frac{\alpha + \beta}{\alpha \beta} =-3}$