(B) Given the equation $f(x) = ax^3 + ax^2 + bx + c = 0$. Since $x = -1$ is a twice repeated root,$f(-1) = 0$ and $f'(-1) = 0$. First,$f(-1) = a(-1)^3

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(B) Given the equation $f(x) = ax^3 + ax^2 + bx + c = 0$. Since $x = -1$ is a twice repeated root,$f(-1) = 0$ and $f'(-1) = 0$. First,$f(-1) = a(-1)^3 + a(-1)^2 + b(-1) + c = -a + a - b + c = 0$,which implies $c = b$. Next,find the derivative: $f'(x) = 3ax^2 + 2ax + b$. Setting $f'(-1) = 0$: $3a(-1)^2 + 2a(-1) + b = 3a - 2a + b = a + b = 0$. This gives $a = -b$. Substituting these into the ratio: $a:b:c = (-b):b:b = -1:1:1$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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If $-1$ is a twice repeated root of the equation $a(x^3+x^2)+bx+c=0$,then $a:b:c=$

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A
$1:-1:1$
B
$-1:1:1$
C
$1:1:-1$
D
$1:1:1$

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If $-1$ is a twice repeated root of the equation $a(x^3+x^2)+bx+c=0$,then $a:b:c=$

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