If the roots of x^5-ax^4+bx^3-cx^2+dx-1=0 are | vedclass

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(D) Let the roots of the equation be $x_1, x_2, x_3, x_4, x_5$. Since all roots are positive and their arithmetic mean $(AM)$ equals their geometric m

Vedclass · Accepted Answer

$30$

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(D) Let the roots of the equation be $x_1, x_2, x_3, x_4, x_5$. Since all roots are positive and their arithmetic mean $(AM)$ equals their geometric mean $(GM)$,all roots must be equal. Let $x_1 = x_2 = x_3 = x_4 = x_5 = \alpha$. From the equation $x^5-ax^4+bx^3-cx^2+dx-1=0$,the product of the roots is $x_1 x_2 x_3 x_4 x_5 = (-1)^5 (-1) = 1$. Thus,$\alpha^5 = 1$,which implies $\alpha = 1$. The equation is $(x-1)^5 = x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1 = 0$. Comparing this with the given equation $x^5-ax^4+bx^3-cx^2+dx-1=0$,we get $a=5, b=10, c=10, d=5$. Therefore,$a+b+c+d = 5+10+10+5 = 30$.

If the roots of $x^5-ax^4+bx^3-cx^2+dx-1=0$ are all positive such that their arithmetic mean and geometric mean are equal,then $a+b+c+d=$

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