Let - be the roots of quadratic equation ax2-bx-c-0- If 1- are in arithmetic progression and 1- 12-1- are also in arithmetic progression- then the value of -2-2-2-2-2-2-2 is

1,α+β,αβ are in A.P., so 1+αβ=2(α+β) ⇒α+β=1+αβ2 ⋯(1) Also, 1α,12,1β are in A.P., so 1α+1β=1 ⇒α+β=αβ ⋯(2) From equation (1) and (2), we get α+β=αβ=1 ∴α^2+β ...

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Standard XII

Mathematics

Common Roots

Let α,β be th...

Question

Let $α, β$ be the roots of quadratic equation $a x^{2} + b x + c = 0.$ If $1, α + β, α β$ are in arithmetic progression and $\frac{1}{α}, \frac{1}{2}, \frac{1}{β}$ are also in arithmetic progression, then the value of $\frac{α^{2} + β^{2} - 2 α^{2} β^{2}}{α^{2} + β^{2}}$ is

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Solution

The correct option is B $3$
$1, α + β, α β$ are in A.P., so
$1 + α β = 2 (α + β)$
$\Rightarrow α + β = \frac{1 + α β}{2} \dots (1)$

Also, $\frac{1}{α}, \frac{1}{2}, \frac{1}{β}$ are in A.P., so
$\frac{1}{α} + \frac{1}{β} = 1$
$\Rightarrow α + β = α β \dots (2)$
From equation $(1)$ and $(2)$ , we get
$α + β = α β = 1$

$∴ \frac{α^{2} + β^{2} - 2 α^{2} β^{2}}{α^{2} + β^{2}}$
$= 1 - \frac{2 α^{2} β^{2}}{α^{2} + β^{2}} = 1 - \frac{2 α^{2} β^{2}}{(α + β)^{2} - 2 α β} = 3$

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Let α,β be the roots of quadratic equation ax2+bx+c=0. If 1,α+β,αβ are in arithmetic progression and 1α,12,1β are also in arithmetic progression, then the value of α2+β2−2α2β2α2+β2 is

The correct option is B 31,α+β,αβ are in A.P., so 1+αβ=2(α+β) ⇒α+β=1+αβ2 ⋯(1) Also, 1α,12,1β are in A.P., so 1α+1β=1 ⇒α+β=αβ ⋯(2) From equation (1) and (2), we get α+β=αβ=1 ∴α2+β2−2α2β2α2+β2 =1−2α2β2α2+β2=1−2α2β2(α+β)2−2αβ=3

Let $α, β$ be the roots of quadratic equation $a x^{2} + b x + c = 0.$ If $1, α + β, α β$ are in arithmetic progression and $\frac{1}{α}, \frac{1}{2}, \frac{1}{β}$ are also in arithmetic progression, then the value of $\frac{α^{2} + β^{2} - 2 α^{2} β^{2}}{α^{2} + β^{2}}$ is