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If one root of ax+bx+c-0 is equal to nth power of theother, then the value of (ac)i + (ac)i is |
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Answer» If one root of a quadratic equation (ax^2+bx+c) is the nth power of another, how do you prove that (a^n*c) + (a*c^n) +b=0?Let the two roots be α and α^n. We know that sum of the roots is (-b/a) and product of the roots is (c/a). (α)(α^n) = c/a -> α^(n+1)= c/a -> α = (c/a)^(1/(n+1)) α + α^n = -b/a (Now let's replace α with the value we got) (c/a)^(1/(n+1)) + (c/a)^(n/(n+1)) = -b/a ((a^n)c)^(1/(n+1)) + (a(c^n))^(1/(n+1)) = -b ((a^n)c)^(1/(n+1)) + (a(c^n))^(1/(n+1)) + b = 0 Therefore, value of ((a^n)c)^(1/(n+1)) + (a(c^n))^(1/(n+1)) = - b |
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