Question

Solve the following equations:
$ \sqrt {2{x^2} - 7x + 1} - \sqrt {2{x^2} - 9x + 4} = 1. \\
$

Answer 1

Hint: Simplify the equation using the normal method by bringing common terms together on one side and then solve to find the answer.

 The given equation is as
\[ \sqrt {2{x^2} - 7x + 1} - \sqrt {2{x^2} - 9x + 4} = 1 \\
   \Rightarrow \sqrt {2{x^2} - 7x + 1} = \sqrt {2{x^2} - 9x + 4} + 1 \\ \]
 After squaring both sides, we get,
  $2{x^2} - 7x + 1 = 1 + 2{x^2} - 9x + 4 + 2\sqrt {2{x^2} - 9x + 4} \\
  2x - 4 = 2\sqrt {2{x^2} - 9x + 4} \\
  x - 2 = \sqrt {2{x^2} - 9x + 4} \\$
 Again after squaring both sides, we get,
  ${(x - 2)^2} = 2{x^2} - 9x + 4 \\
  {x^2} + 4 - 4x = 2{x^2} - 9x + 4 \\
  {x^2} - 5x = 0 \\
  x(x - 5) = 0 \\
   \Rightarrow x = 0,5 \\$
So, this is the required solution.

NOTE: On squaring the values, we must write the terms carefully, without missing any terms in between.