Question

If α and β are the zeros of the polynomial, f(x) = 6x2 + x – 2 , find :
α / β + β / α

Answer 1

Given equation,

$6x^{2} +x-2=0$

Solving, we get,

$6x^{2} -3x+4x-2=0\\3x(2x-1)+2(2x-1)=0\\(3x+2)(2x-1)=0\\\\\bold{Now}\\3x+2=0\\3x=-2\\\boxed{\boxed{\bold{x=\frac{-2}{3}} }}\\\\2x-1=0\\2x=1\\\boxed{\boxed{\bold{x=\frac{1}{2} }}}$

Let alpha and beta be -2/3 and 1/2

$=> \alpha /\beta + \beta /\alpha \\=>\boxed{\boxed{\bold {\frac{-25}{12} }}}$

                                                         

Answer 2

Answer:

Polynomial given: 6x²+x-2

By, splitting the middle term.

➜ 6x²+4x-3x-2

➜ 2x(3x+2)-1(3x+2)

➜(2x-1)(3x+2)

➜ x=1/2,-2/3

Now, let

• α = 1/2
• β = -2/3

Given: α / β + β / α

➜(α² + β²)/αβ

➜(a+ β)²-2aβ/αβ

• Sum of zeroes = α + β = -b/a = -1/6
• Product of zeroes = αβ = c/a = -2/6 = -1/3

➜((-1/6)²-2(-1/3))/(-1/3)

➜((1/36)+2/3)/(-1/3)

➜(1+24/36)/(-1/3)

➜25/36 × -3/1

➜-25/12

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Some formulas:

• Sum of zeroes = -(coefficient of x)/ coefficient of x²
• Product of zeros = coefficient of constant term /coefficient of x²

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