If - - are the roots of the equation 3x2-6x-4-0- the value of -21-1-3- will be

The correct option is D 8 α+β=6/3=2 ...(i) α·β=4/3 ...(ii) (α/β+β/α)+2(1/α+1/β)+3αβ =(α^2+β^2/αβ)+2(α+β/αβ)+3αβ =(α+β)^2 2αβ/αβ+2 ...

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Standard XII

Mathematics

Algebra of Limits

If α, β a...

Question

If $α$ , $β$ are the roots of the equation $3 x^{2} - 6 x + 4 = 0$ , the value of $(\frac{α}{β} + \frac{β}{α}) + 2 (\frac{1}{α} + \frac{1}{β}) + 3 α β$ will be

6

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\frac{4}{3}

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2

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8

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Solution

The correct option is D $8$
$α + β = \frac{6}{3} = 2$ ...(i)
$α \cdot β = \frac{4}{3}$ ...(ii)
$(\frac{α}{β} + \frac{β}{α}) + 2 (\frac{1}{α} + \frac{1}{β}) + 3 α β$
$= (\frac{α^{2} + β^{2}}{α β}) + 2 (\frac{α + β}{α β}) + 3 α β$
$= \frac{(α + β)^{2} - 2 α β}{α β} + 2 (\frac{2 \times 3}{4}) + \frac{3 \times 4}{3}$
$= \frac{4 - \frac{8}{3}}{\frac{4}{3}} + 3 + 4 = 1 + 7 = 8$

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If α, β are the roots of the equation 3x2−6x+4=0, the value of (αβ+βα)+2(1α+1β)+3αβ will be

The correct option is D 8α+β=63=2 ...(i)α⋅β=43 ...(ii)(αβ+βα)+2(1α+1β)+3αβ=(α2+β2αβ)+2(α+βαβ)+3αβ=(α+β)2−2αβαβ+2(2×34)+3×43=4−8343+3+4=1+7=8

If $α$ , $β$ are the roots of the equation $3 x^{2} - 6 x + 4 = 0$ , the value of $(\frac{α}{β} + \frac{β}{α}) + 2 (\frac{1}{α} + \frac{1}{β}) + 3 α β$ will be