If α,β and γ are the zeroes of the polynomial f(x)=ax3+bx2+cx+d, then 1α+1β+1γ i
Question
If α,β and γ are the zeroes of the polynomial f(x)=ax3+bx2+cx+d, then 1α+1β+1γ i
Solution
The sum of the reciprocals of the roots of a cubic equation is equal to the ratio of the coefficient of the second term to the coefficient of the third term.
Given the cubic equation f(x) = ax^3 + bx^2 + cx + d, where α, β, and γ are the roots, the sum of the reciprocals of the roots is given by:
1/α + 1/β + 1/γ = c/a
This is derived from Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots.
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For example if α, β, γ are roots of ax3 + bx2 + cx + d = 0 to find equation whose roots are 1/α, 1/β, 1/γ, we put 1α = y ⇒ α = 1y . As α is a root of ax3 + bx2 + cx + d = 0, we get ay3 + by2 + cy + d = 0 or a + by + cy2 + dy3 = 0 This is the desired equation.The same result holds for all polynomial equations.If α, β are roots of ax2 + bx + c = 0 the roots of a(x − 1)2 + b(x − 1)(x − 2) + c(x − 2)2 = 0 are
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