Solving Radical Equations x−7=x−5

Step 1
We can avoid squaring both sides.
Let
${\displaystyle x-5=u^2}$,
where
${\displaystyle u\ge 0}$
Then
${\displaystyle \sqrt{x-5}=u}$
Also,
${\displaystyle x=u^2+5}$
so
${\displaystyle x-7=u^2-2}$
Thus our equation can be rewritten as
${\displaystyle u^2-2=u}$
or equivalently
${\displaystyle u^2-u-2=0}$
But
${\displaystyle u^2-u-2=(u-2)(u+1)}$
Thus the solutions of
${\displaystyle u^2-u-2=0}$
are
${\displaystyle u=2}$
and  ${\displaystyle u=-1}$
Since ${\displaystyle u\ge 0}$, we reject the solution ${\displaystyle u=-1}$
We conclude that ${\displaystyle u=2}$, and therefore ${\displaystyle x=u^2+5=9}$.

Step 1
${\displaystyle x-7=\sqrt{x-5}}$
${\displaystyle {(x-7)}^2={\sqrt{x-5}}^2}$
${\displaystyle {(x-7)}^2=x-5}$
${\displaystyle (x-7)(x-7)=x-5}$
${\displaystyle x^2-7x-7x+49=x-5}$
${\displaystyle x^2-15x+54=0}$
${\displaystyle (x-9)(x-6)=0}$
${\displaystyle x-9=0}$ or ${\displaystyle x-6=0}$
${\displaystyle x=9}$ or ${\displaystyle x=6}$
Now check for extraneous solutions

Step 1 $(x-7)=\sqrt{x-5}$ Check the domain first $x-5\ge 0\cap x-7\ge 0$ $x\ge 7$ $(x-7{)}^2={\sqrt{x-5}}^2$ $(x-7{)}^2=x-5$ $(x-7)(x-7)=x-5$ $x^2-7x-7x+49=x-5$ $x^2-15x+54=0$ $(x-9)(x-6)=0$ $x-9=0$ or $x-6=0$ $x=9$ or $x=6$ checking the domain $x=6$ is an extraneous root. $x=9$