Two roots of the equation ax^4 + bx^3 + cx^2 | vedclass

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(B) Let the roots be $\alpha, \alpha, \beta, \gamma$. Given $\alpha > 0$ and $\beta\gamma = 1$. From Vieta's formulas,the product of roots is $\alpha^

Vedclass · Accepted Answer

$3e + \frac{2b\sqrt{e}}{\sqrt{a}} + c = a$

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(B) Let the roots be $\alpha, \alpha, \beta, \gamma$. Given $\alpha > 0$ and $\beta\gamma = 1$. From Vieta's formulas,the product of roots is $\alpha^2\beta\gamma = \frac{e}{a}$. Since $\beta\gamma = 1$,we have $\alpha^2 = \frac{e}{a}$,so $\alpha = \sqrt{\frac{e}{a}}$. The equation can be written as $a(x-\alpha)^2(x^2 - Sx + 1) = 0$,where $S = \beta + \gamma$. Expanding $a(x^2 - 2\alpha x + \alpha^2)(x^2 - Sx + 1) = a(x^4 - (S+2\alpha)x^3 + (1 + 2\alpha S + \alpha^2)x^2 - (S\alpha^2 + 2\alpha)x + \alpha^2) = 0$. Comparing coefficients with $ax^4 + bx^3 + cx^2 + dx + e = 0$: $b = -a(S + 2\alpha)$,$c = a(1 + 2\alpha S + \alpha^2)$,$d = -a(S\alpha^2 + 2\alpha)$,$e = a\alpha^2$. From $b = -a(S + 2\alpha)$,we get $S = -\frac{b}{a} - 2\alpha$. Substitute $S$ into the expression for $c$: $c = a(1 + 2\alpha(-\frac{b}{a} - 2\alpha) + \alpha^2) = a(1 - \frac{2b\alpha}{a} - 4\alpha^2 + \alpha^2) = a - 2b\alpha - 3a\alpha^2$. Since $\alpha^2 = \frac{e}{a}$,then $3a\alpha^2 = 3e$. Thus,$c = a - 2b\sqrt{\frac{e}{a}} - 3e$,which rearranges to $3e + \frac{2b\sqrt{e}}{\sqrt{a}} + c = a$.

Two roots of the equation $ax^4 + bx^3 + cx^2 + dx + e = 0$ are positive and equal. If the product of the other two real roots is $1$,then:

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