KeerthanaPosted on If αalphaα and βetaβ are the roots of the equation x2−x+3x^{2}-x+3x2−x+3 =0=0=0, then what is the value of (α4+β4)?left(alpha^{4}+eta^{4} ight) ?(α4+β4)? a9 b11 c7 d13 Answer : Option CExplanation : If the roots of quadratic equation ax2 + bx + c = 0 be D and Ethen α+β=−ba;αβ=caalpha+eta=-frac{b}{a} ; alpha eta=frac{c}{a}α+β=−ab;αβ=ac ∴ herefore∴ For equation x2−x+3=0x^{2}-x+3=0x2−x+3=0, α+β=1;αβ=3alpha+eta=1 ; alpha eta=3α+β=1;αβ=3 ∴α4+β4=(α2)2+(β2)2=(α2+β2)2−2α2β2 herefore alpha^{4}+eta^{4}=left(alpha^{2} ight)^{2}+left(eta^{2} ight)^{2}=left(alpha^{2}+eta^{2} ight)^{2}-2 alpha^{2} eta^{2}∴α4+β4=(α2)2+(β2)2=(α2+β2)2−2α2β2 ={(a+β)2−2αβ}2−2α2β2=left{(a+eta)^{2}-2 alpha eta ight}^{2}-2 alpha^{2} eta^{2}={(a+β)2−2αβ}2−2α2β2 =(1−2×3)2−2×9=(−5)2−18=25−18=7=(1-2 imes 3)^{2}-2 imes 9=(-5)^{2}-18=25-18=7=(1−2×3)2−2×9=(−5)2−18=25−18=7 Rate This:NaN / 5 - 1 votesAdd comment