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

Let | x-1|= t t^2-5 t+6=0 \ t=2 \ t=3 \ | x-1|=2 \ | x-1|=3 \ x-1 ± 2 \ x-1= ± 3 \ x=1 ± 2 \ x=1 ± 3 \ Therefore root =3,-1,4,-2 \ Therefore Sum of root =3+(-1)+4+(-2)=4

Let c and d be vectors such that c d sqrt29 and c x 2 i... | null (Mathematics)

Let

|\mathrm{x}-1|=\mathrm{t}\\

\begin{aligned} & \mathrm{t}^{2}-5 \mathrm{t}+6=0 \\ & \mathrm{t}=2 \& \mathrm{t}=3 \\ & |\mathrm{x}-1|=2 \&|\mathrm{x}-1|=3 \\ & \mathrm{x}-1 \pm 2 \& \mathrm{x}-1= \pm 3 \\ & \mathrm{x}=1 \pm 2 \& \mathrm{x}=1 \pm 3 \\ & \therefore \text { root }=3,-1,4,-2 \\ & \therefore \text { Sum of root }=3+(-1)+4+(-2)=4 \end{aligned}

Explanation

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