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Question Number 140973 by EnterUsername last updated on 14/May/21
If α and β are roots of the equation 2x^2 +ax+b=0,  then one of the roots of the equation 2(αx+β)^2 +  a(αx+β)+b=0 is  (A) 0                                       (B) ((α+2b)/α^2 )  (C) ((aα+b)/(2α^2 ))                              (D) ((aα−2b)/(2α^2 ))
Ifαandβarerootsoftheequation2x2+ax+b=0,thenoneoftherootsoftheequation2(αx+β)2+a(αx+β)+b=0is(A)0(B)α+2bα2(C)aα+b2α2(D)aα2b2α2
Answered by TheSupreme last updated on 15/May/21
α=((−a+(√Δ))/4)  β=((−a−(√Δ))/4)  2(αx+β)^2 +a(αx+β)+b=0  αx+β=α  x=((α−β)/α)  αx+β=β  x=0 (A)
α=a+Δ4β=aΔ42(αx+β)2+a(αx+β)+b=0αx+β=αx=αβααx+β=βx=0(A)
Commented by EnterUsername last updated on 19/May/21
Thanks so much
Thankssomuch

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