Roots

we get by the answer just by rearranging $r^2-6r+5=0$ $\Rightarrow(r^2-6r+9)-9+5=0$ $\Rightarrow(r-3)^2=4$

${ r }^{ 2 }\quad -\quad 6r\quad +\quad 5\quad =\quad 0\quad \\ by\quad factorization\quad method,\\ { r }^{ 2 }-\quad 5r\quad -\quad 1r\quad +\quad 5\quad =\quad 0\\ \Rightarrow \quad r({ r }-\quad 5)\quad -1(r\quad -\quad 5)\quad =\quad 0\\ \Rightarrow \quad (r-1)(r-5)\quad =\quad 0\\ \\ r-1=0\quad \quad \quad \quad or\quad \quad r-5=0\\ r=1\quad \quad \quad \quad \quad \quad or\quad \quad r=5\\ \\ Case\quad I\quad \quad \quad \quad \quad \quad \quad Case\quad II\\ r=1\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad r=5\\ { (r-3) }^{ 2 }\quad \quad \quad \quad \quad \quad \quad \quad { (r-3) }^{ 2 }\\ ={ (1-3) }^{ 2 }\quad \quad \quad \quad \quad \quad { =(5-3) }^{ 2 }\\ ={ (-2) }^{ 2 }\quad \quad \quad \quad \quad \quad \quad ={ (2) }^{ 2 }\\ =\boxed { 4 } \quad \quad \quad \quad \quad \quad \quad \quad \quad =\boxed { 4 }$

THe factor of r^2 -6r +5 is (r-5) (r-1), assume that r=5 substituting to the given equation 5^2 - 6(5) = -5 or 25-30 = -5 thus, -5 = -5 is correct so the value of r is 5 ... then substituting to the another equation ((r-3)^2 = (5-3)^2 is equal to 4.

Whatever.. that is how i solve it hope u understand!

When we factorise $r^{2}$ - 6r + 5 = 0, we get factors as

(x-1) & (x-5)

When we equate any one of the factor to 0, we get

x-5=0 $\Rightarrow$ x=5

By substituting 5 in $(r-3)^{2}$ , we get

$(5-3)^{2}$ $\Rightarrow$ $2^{2}$ =4

Therefore the value of $(r-3)^{2}$ is $\boxed{4}$

$\color{cerulean}{r^2-6r+5=(r^2-6r+9)-4=0\rightarrow (r-3)^2=\boxed{4}}$

${ r }^{ 2 }-6r+5=0.........(A)\\ Adding\quad "4"\quad on\quad both\quad sides\quad of\quad (A)\\ =>{ r }^{ 2 }-6r+9=4\\ =>\boxed { { (r-3) }^{ 2 }=4 }$

solve the equation and we get r=5 , r=1 substituting both values in the given expression we get 4