If - and Y are zeros of 6 x 3-3 x 2-5 x -1 then find 1-1- and 1-

Α +β +γ = 3/6 α +β +γ = 1/2 (α×β)+(α×γ)+(β×γ)= 5/6 (α×β×γ)= 1/6 1/α+1/β+1/γ = ( β×γ/α×β×γ )+ ( α×γ/α×β×γ )+ ( α×β/α×β×γ ) ∴,= [ {α×β} + {α×γ} +{β×γ}/{α×β×γ} ] ...

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Byju's Answer

Standard X

Mathematics

Relationship Between Zeroes and Coefficients of a Quadratic Polynomial

If αβ and Y a...

Question

If $α β$ and $γ$ are zeros of $6 x^{3} + 3 x^{2} - 5 x + 1$ then find $\frac{1}{α} + \frac{1}{β}$ and $\frac{1}{γ}$ .

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Solution

$α + β + γ = - \frac{3}{6} α + β + γ = - \frac{1}{2} (α \times β) + (α \times γ) + (β \times γ) = - \frac{5}{6} (α \times β \times γ) = - \frac{1}{6} \frac{1}{α} + \frac{1}{β} + \frac{1}{γ}$
$= (\frac{β \times γ}{α \times β \times γ}) + (\frac{α \times γ}{α \times β \times γ}) + (\frac{α \times β}{α \times β \times γ}) ∴, = [\frac{{α \times β} + {α \times γ} + {β \times γ}}{{α \times β \times γ}}]$
putting values
$= \frac{(- \frac{5}{6})}{(- \frac{1}{6})} = \frac{- \frac{5}{6}}{- \frac{1}{6}} = \frac{- 5 \times 6}{- 1 \times 6} = \frac{5}{1} = 5 ∴ \frac{1}{α} + \frac{1}{β} + \frac{1}{γ} = 5$

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If αβ and γ are zeros of 6x3+3x2−5x+1 then find 1α+1β and 1γ.

α+β+γ=−36α+β+γ=−12(α×β)+(α×γ)+(β×γ)=−56(α×β×γ)=−161α+1β+1γ =(β×γα×β×γ)+(α×γα×β×γ)+(α×βα×β×γ)∴,=[{α×β}+{α×γ}+{β×γ}{α×β×γ}] putting values =(−56)(−16)=−56−16=−5×6−1×6=51=5∴1α+1β+1γ=5

If $α β$ and $γ$ are zeros of $6 x^{3} + 3 x^{2} - 5 x + 1$ then find $\frac{1}{α} + \frac{1}{β}$ and $\frac{1}{γ}$ .