(A) Let $|x-1|=t$. Then the equation becomes $t^{2}-5t+6=0$. Factoring the quadratic equation, we get $(t-2)(t-3)=0$, so $t=2$ or $t=3$. Case $1$: $|x

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(A) Let $|x-1|=t$. Then the equation becomes $t^{2}-5t+6=0$. Factoring the quadratic equation, we get $(t-2)(t-3)=0$, so $t=2$ or $t=3$. Case $1$: $|x-1|=2 \implies x-1=2$ or $x-1=-2$, which gives $x=3$ or $x=-1$. Case $2$: $|x-1|=3 \implies x-1=3$ or $x-1=-3$, which gives $x=4$ or $x=-2$. The roots are $3, -1, 4, -2$. The sum of the roots is $3 + (-1) + 4 + (-2) = 4$.

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

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The sum of all the roots of the equation $(x-1)^{2}-5|x-1|+6=0$ is:

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A
$4$
B
$3$
C
$1$
D
$5$

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