If - - - are the zeros of 6x2-x-2 - then - - -

The correct option is A 25/12Sum of zeros (α+β) = 1/6 Product of zeros αβ = nbsp; 1/3 α/β+β/α = α^2+β^2/αβ =α+β2 2αβαβ nbsp; α/β+β/α = nbsp;162 2 13 1 ...

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Byju's Answer

Standard X

Mathematics

Relationship between Zeroes and Coefficients of a Polynomial

If α & β are ...

Question

If $α & β$ are the zeros of $6 x^{2} - x - 2$ , then $\frac{α}{β} + \frac{β}{α} = ?$

\frac{- 25}{12}

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\frac{- 12}{25}

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\frac{24}{25}

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\frac{25}{24}

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Solution

The correct option is A $\frac{- 25}{12}$
Sum of zeros ( $α + β$ ) = $\frac{1}{6}$
Product of zeros $α β$ = $\frac{- 1}{3}$

$\frac{α}{β} + \frac{β}{α} = \frac{α^{2} + β^{2}}{α β}$ = $\frac{{(α + β)}^{2} - 2 αβ}{αβ}$
$\frac{α}{β} + \frac{β}{α} =$ $\frac{{(\frac{1}{6})}^{2} - 2 (\frac{- 1}{3})}{\frac{- 1}{3}}$ = $\frac{\frac{1}{36} + \frac{2}{3}}{\frac{- 1}{3}} = \frac{\frac{25}{36}}{\frac{- 1}{3}}$
= $\frac{- 25}{12}$

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Similar questions

If $α$ and $β$ are the zeros of a polynomial
$f (x) = 6 x^{2} + x - 2$ , find the values of $\frac{α}{β} + \frac{β}{α}$ .

If $α$ and $β$ are the zeros of the polynomial $f (x) = 6 x^{2} + x - 2$ , find the value of $(\frac{α}{β} + \frac{β}{α}) .$

Q. If

α

and

β

are the zeros of the quadratic polynomial

f (x) = 6 x^{2} + x - 2

, find the value of

\frac{α}{β} + \frac{β}{α}

Q. If

α

and

β

are the zeroes of the equation

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

, find

\frac{α}{β} + \frac{β}{α}

Q. If the equation

{(3 - {log}_{\sqrt[12]{4}} (x - 2))}_{\sqrt[12]{4}}^{2} - 4 ∣ ∣ 3 - {log}_{\sqrt[12]{4}} (x - 2) ∣ ∣ + 3 = 0

has integral roots

α

and

β

such that

| α | + | β - 1 | = | α - 1 | + | β |,

then

| α + β |

is equal to

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If α & β are the zeros of 6x2−x−2 , then αβ+βα=?

The correct option is A −2512Sum of zeros (α+β) = 16 Product of zeros αβ = −13 αβ+βα=α2+β2αβ =(α+β)2−2αβαβ αβ+βα= (16)2−2(−13)−13= 136+23−13=2536−13 =−2512

If $α & β$ are the zeros of $6 x^{2} - x - 2$ , then $\frac{α}{β} + \frac{β}{α} = ?$