(B) Given the cubic equation: $x^3-6x^2+6x-5=0$. To transform the roots by increasing them by $h$,we replace $x$ with $(x+h)$. The new equation become

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(B) Given the cubic equation: $x^3-6x^2+6x-5=0$. To transform the roots by increasing them by $h$,we replace $x$ with $(x+h)$. The new equation becomes: $(x+h)^3-6(x+h)^2+6(x+h)-5=0$. Expanding the terms: $(x^3+3x^2h+3xh^2+h^3)-6(x^2+2xh+h^2)+6(x+h)-5=0$. Grouping the $x^2$ terms: $x^3 + (3h-6)x^2 + (3h^2-12h+6)x + (h^3-6h^2+6h-5) = 0$. Since the new equation does not contain the $x^2$ term,the coefficient of $x^2$ must be zero: $3h-6=0$. Solving for $h$: $3h=6 \Rightarrow h=2$.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Solution of quadratic equations and Nature of roots

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Each of the roots of the equation $x^3-6x^2+6x-5=0$ are increased by $h$. If the new transformed equation does not contain the $x^2$ term,then $h$ is equal to:

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$1$
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$2$
C
$\frac{1}{2}$
D
$\frac{1}{3}$

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Each of the roots of the equation $x^3-6x^2+6x-5=0$ are increased by $h$. If the new transformed equation does not contain the $x^2$ term,then $h$ is equal to:

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