Assume that alpha, beta, gamma are the roots | vedclass

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(A) The roots of the equation $2x^3+5x^2+5x+2=0$ are $\alpha, \beta, \gamma$. From Vieta's formulas:$\alpha+\beta+\gamma = -\frac{5}{2}$$\alpha\beta+\

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$c(h) 
eq 0, \forall h \in R$

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(A) The roots of the equation $2x^3+5x^2+5x+2=0$ are $\alpha, \beta, \gamma$. From Vieta's formulas:$\alpha+\beta+\gamma = -\frac{5}{2}$$\alpha\beta+\beta\gamma+\gamma\alpha = \frac{5}{2}$$\alpha\beta\gamma = -1$Let the new roots be $y = x+h$,so $x = y-h$. Substituting into the original equation:$2(y-h)^3+5(y-h)^2+5(y-h)+2 = 0$$2(y^3-3y^2h+3yh^2-h^3)+5(y^2-2yh+h^2)+5(y-h)+2 = 0$$2y^3 + (5-6h)y^2 + (6h^2-10h+5)y + (-2h^3+5h^2-5h+2) = 0$Comparing this with $a(h)x^3+b(h)x^2+c(h)x+d(h)=0$,we have $a(h)=2$,$b(h)=5-6h$,$c(h)=6h^2-10h+5$,and $d(h)=-2h^3+5h^2-5h+2$.For $c(h) = 6h^2-10h+5$,the discriminant $D = (-10)^2 - 4(6)(5) = 100 - 120 = -20 Since the discriminant is negative and the leading coefficient is positive,$c(h) > 0$ for all $h \in R$. Thus,$c(h) 
eq 0$ for all $h \in R$.

Assume that $\alpha, \beta, \gamma$ are the roots of $2x^3+5x^2+5x+2=0$. For $h \in R$,if $\alpha+h, \beta+h, \gamma+h$ are roots of $a(h)x^3+b(h)x^2+c(h)x+d(h)=0$,then:

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