If -1 is a twice repeated root of the equatio | vedclass

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(C) Let the roots of the cubic equation $ax^3+bx^2+cx+1=0$ be $-1, -1, \alpha$. From the relation between roots and coefficients,the product of roots

Vedclass · Accepted Answer

$b=2a+1, c=a+2$

Vedclass · Answer

(C) Let the roots of the cubic equation $ax^3+bx^2+cx+1=0$ be $-1, -1, \alpha$. From the relation between roots and coefficients,the product of roots is $\alpha 	imes (-1) 	imes (-1) = -\frac{1}{a}$,which implies $\alpha = -\frac{1}{a}$. The sum of roots is $(-1) + (-1) + \alpha = -\frac{b}{a}$. Substituting $\alpha = -\frac{1}{a}$,we get $-2 - \frac{1}{a} = -\frac{b}{a}$. Multiplying by $-a$,we get $2a + 1 = b$,so $b = 2a + 1$. Since $-1$ is a root,$a(-1)^3 + b(-1)^2 + c(-1) + 1 = 0$,which simplifies to $-a + b - c + 1 = 0$. Substituting $b = 2a + 1$,we get $-a + (2a + 1) - c + 1 = 0$,which simplifies to $a + 2 - c = 0$,hence $c = a + 2$.

If $-1$ is a twice repeated root of the equation $ax^3+bx^2+cx+1=0$,then

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