If alpha, beta, gamma are the roots of the eq | vedclass

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(A) Let the roots of $x^3-5x^2+4x=0$ be $y_1, y_2, y_3$. Factoring the equation,we get $x(x^2-5x+4)=0$,so $x(x-1)(x-4)=0$. The roots are $0, 1, 4$. Gi

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$5$

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(A) Let the roots of $x^3-5x^2+4x=0$ be $y_1, y_2, y_3$. Factoring the equation,we get $x(x^2-5x+4)=0$,so $x(x-1)(x-4)=0$. The roots are $0, 1, 4$. Given $(\alpha-2)^2, (\beta-2)^2, (\gamma-2)^2$ are $0, 1, 4$. Thus,$(\alpha-2)^2=0 \implies \alpha=2$. $(\beta-2)^2=1 \implies \beta-2 = \pm 1 \implies \beta=3$ or $\beta=1$. $(\gamma-2)^2=4 \implies \gamma-2 = \pm 2 \implies \gamma=4$ or $\gamma=0$. By Vieta's formulas for $x^3-Px^2+Qx-R=0$: $P = \alpha+\beta+\gamma$,$Q = \alpha\beta+\beta\gamma+\gamma\alpha$,$R = \alpha\beta\gamma$. We want to minimize $P+Q+R$. Note that $P+Q+R = (\alpha+1)(\beta+1)(\gamma+1)-1$. Testing combinations: If $\alpha=2, \beta=3, \gamma=4$,then $P=9, Q=26, R=24$,$P+Q+R=59$. If $\alpha=2, \beta=1, \gamma=0$,then $P=3, Q=2, R=0$,$P+Q+R=5$. If $\alpha=2, \beta=3, \gamma=0$,then $P=5, Q=6, R=0$,$P+Q+R=11$. If $\alpha=2, \beta=1, \gamma=4$,then $P=7, Q=14, R=8$,$P+Q+R=29$. Considering negative roots for $\beta-2$ and $\gamma-2$: If $\beta=1, \gamma=0$,$P=3, Q=2, R=0$,sum $= 5$. If $\beta=3, \gamma=0$,$P=5, Q=6, R=0$,sum $= 11$. If $\beta=1, \gamma=4$,$P=7, Q=14, R=8$,sum $= 29$. If $\beta=3, \gamma=4$,$P=9, Q=26, R=24$,sum $= 59$. The least value is $5$.

List-$I$	List-$II$
$A$. All the roots are negative	$I$. $(b - 3)^2 = 36 + P^2$ for $P \in R$
$B$. Two roots are complex	$II$. $-3 < b < 9$
$C$. Two roots are positive	$III$. $b \in (-\infty, -3) \cup (9, \infty)$
$D$. All roots are real and distinct	$IV$. $b = 9$
	$V$. $b = -3$

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-Px^2+Qx-R=0$ and $(\alpha-2)^2, (\beta-2)^2, (\gamma-2)^2$ are the roots of the equation $x^3-5x^2+4x=0$,then the possible least value of $P+Q+R$ is

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