With respect to the roots of the equation 3x^ | vedclass

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(A) The given equation is $3x^3 + bx^2 + bx + 3 = 0$.Factoring the equation: $3(x^3 + 1) + bx(x + 1) = 0$.$3(x + 1)(x^2 - x + 1) + bx(x + 1) = 0$.$(x

Vedclass · Accepted Answer

$A-IV, B-II, C-V, D-III$

Vedclass · Answer

(A) The given equation is $3x^3 + bx^2 + bx + 3 = 0$.Factoring the equation: $3(x^3 + 1) + bx(x + 1) = 0$.$3(x + 1)(x^2 - x + 1) + bx(x + 1) = 0$.$(x + 1)(3x^2 - 3x + 3 + bx) = 0$.$(x + 1)(3x^2 + (b - 3)x + 3) = 0$.One root is $x = -1$. The other two roots are roots of $3x^2 + (b - 3)x + 3 = 0$.Let $f(x) = 3x^2 + (b - 3)x + 3$.For $A$: All roots are negative. If $b = 9$,$f(x) = 3x^2 + 6x + 3 = 3(x + 1)^2$. Roots are $-1, -1, -1$. All negative. So $A ightarrow IV$.For $B$: Two roots are complex. Discriminant $D For $C$: Two roots are positive. If $b = -3$,$f(x) = 3x^2 - 6x + 3 = 3(x - 1)^2$. Roots are $-1, 1, 1$. Two positive. So $C ightarrow V$.For $D$: All roots are real and distinct. $D > 0$ and $f(-1) 
eq 0$. $D = (b - 3)^2 - 36 > 0 \Rightarrow b \in (-\infty, -3) \cup (9, \infty)$. Also $f(-1) = 3 - (b - 3) + 3 = 9 - b 
eq 0 \Rightarrow b 
eq 9$. So $D ightarrow III$.Thus,$A-IV, B-II, C-V, D-III$.

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	$IV$. $b = 9$
	$V$. $b = -3$

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