Question 728593

{{{sqrt(2x - 3) = x - 1}}}....square both sides

{{{(sqrt(2x - 3))^2 = (x - 1)^2}}}

{{{2x - 3 = x^2-2x +1}}}

{{{0= x^2-2x-2x+3 +1}}}

{{{0= x^2-4x+4}}}...use quadratic formula

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{x = (-(-4) +- sqrt( (-4)^2-4*1*4 ))/(2*1) }}} 

{{{x = (4 +- sqrt( 16-16 ))/2 }}}

 {{{x = (4 +- sqrt( 0 ))/2 }}}

{{{x = (4 +- 0)/2 }}}

so, solution is

{{{x = 4/2 }}}

{{{x = 2 }}}


choice: (c.) 2



{{{sqrt(3x-3) - sqrt(x) = 1}}}

{{{sqrt(3x-3)  = sqrt(x)+1}}}...square both sides

{{{(sqrt(3x-3))^2 = (sqrt(x)+1)^2}}}

{{{3x-3 =x+2sqrt(x)+1}}}

{{{3x-x-3-1 =2sqrt(x)}}}

{{{2x-4 =2sqrt(x)}}}

{{{2(x-2) =2sqrt(x)}}}

{{{cross(2)(x-2) =cross(2)sqrt(x)}}}

{{{x-2 =sqrt(x)}}}..square both sides again

{{{(x-2)^2 =(sqrt(x))^2}}}

{{{x^2-4x+4 =x}}}


{{{x^2-4x-x+4 =0}}}

{{{x^2-5x+4 =0}}}


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-(-5) +- sqrt((-5)^2-4*1*4 ))/(2*1) }}}


{{{x = (5 +- sqrt(25-16 ))/2 }}}

{{{x = (5 +- sqrt(9 ))/2 }}}


{{{x = (5 +- 3)/2 }}}

solutions:

{{{x = (5 +3)/2 }}}

{{{x = 8/2 }}}

{{{x = 4 }}}

and

{{{x = (5 -3)/2 }}}

{{{x = 2/2 }}}

{{{x = 1 }}}


choice:  (c.) 1 and 4