Given α and β are the roots of the equation,7x2−3x−2=0, then α+β=73;αβ=−72
Now
1−α2α+1−β2β=(1−α2)(1−β2)(α+β)−αβ(α+β)=1+(αβ)2−(α2+β2)(α+β)−αβ(α+β)
⇒1+(αβ)2−(α+β)2+2αβ(α+β)−αβ(α+β)=1+(72)2−(73)2−2(72)73+72(73)=1627
If α and β are the roots of the equation, 7x2−3x−2=0, then the value of1−α2α+1−β2β is equal to:
Held on 5 Sept 2020 · Verified 6 Jul 2026.
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1627
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