Solution - Radical equations

Radical Equation entered :

     2y-3+√y+1 = 0

Step by step solution :

Step  1  :

Isolate the square root on the left hand side :

     Original equation
     2y-3+√y+1 = 0

     Isolate
     √y = -2y+3-1+0

      Tidy up
     √y = -2y+2

Step  2  :

Eliminate the radical on the left hand side :

     Raise both sides to the second power
     (√y)² = (-2y+2)²

     After squaring
     y = 4y²-8y+4

Step  3  :

Solve the quadratic equation :

     Rearranged equation
      4y²  - 9y  + 4 = 0

     This equation has two real roots:
       y1 = ( 9 + sqrt( 17) ) / 8 = 1.6404
      y2 = ( 9 - sqrt( 17) ) / 8 = 0.6096

Step  4  :

Check that the first solution is correct :

     Original equation, root isolated, after tidy up
     √y = -2y+2

     Plug in 1.6404 for  y 
      √( 1.6404)  = -2•( 1.6404) +2

      Simplify
      √1.64 = -1.281
      Solution does not check
      1.281 ≠ -1.281 

Step  5  :

Check that the second solution is correct :

     Original equation, root isolated, after tidy up
     √y = -2y+2

     Plug in 0.6096 for  y 
      √( 0.6096)  = -2•( 0.6096) +2

      Simplify
      √0.61 = 0.781
      Solution checks !!
     Solution is:
      y =  0.6096

One solution was found :