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\[\alpha\] is a root of \[4x^{2}+2x-1=0\] and \[f(x)=4x^{3}-3x+1\]then \[2(f(\alpha)+\alpha)=?\]
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\( 4a^2+2a-1=0 \) so u have \( 4a^2=-2a+1 \) then \( 2(f(a)+a)=2f(a)+2a=8a^3-6a+2+2a=8a^3-4a+2=2a(-2a+1)-4a+2 \)
check it
that is f(alpha)
yes i mean alpha by a
ok
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then
well then \(2a(-2a+1)-4a+2=-4a^2-2a+2=-(4a^2+2a-1)+1=0+1=1 \)
let me know if u have any doubt
explain after this 8a3−4a+2
ok 8a^3=4a^2 (2a) and i replaced 4a^2 with -2a+1 because quadratic equation gives us 4a^2=-2a+1
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what about +2 in 8a3−4a+2
2f(a)=2(4a^3-3a+1)=8a^3-6a+2 -----
where is +2 here 8a^3=4a^2 (2a)
sorry which +2
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