Question

how to remove roots...basics/for removing roots what should we ensure...

Answer 1

Here, RHS is completely under root and LHS is a number. Easiest way to solve this equation is to remove the root first. For that, square both sides and solve from there.

$5 = \sqrt{ {(3 - x)}^{2} + ( {2 + 1)}^{2} } \\ \\ {(5)}^{2} = {( \: \: \sqrt{( {3 - x) }^{2} + {(2 + 1)}^{2} } \: \: )}^{2}$

$25 = ( {3 - x) }^{2} + {(2 + 1)}^{2} \\ \\ 25 = (9 + {x}^{2} - 6x) + 9$

$25 = {x}^{2} - 6x+ 9 + 9 \\ \\ {x}^{2} - 6x + 18 - 25 = 0$

${x}^{2} - 6x - 7 = 0 \\ \\ {x}^{2} - 7x + x - 7 = 0 \\ \\ x(x - 7) + 1(x - 7) = 0$

$(x - 7)(x + 1) = 0 \\ \\ x - 7 = 0 \: \: and \: \: x + 1 = 0 \\ \\ x = 7 \: \: and \: \: x = - 1$


Thus x = 7 and -1