Question

Solve the following equations:

$\sqrt{2x-4}-\sqrt{x+5}=1$

Answer 1

Squaring both sides, we get

${(\sqrt{2x-4}-\sqrt{x+5})}^2=1$ which is of the form ${(a-b)}^2$

$\Rightarrow {\left(\sqrt{2x-4}\right)}^2+{\left(\sqrt{x+5}\right)}^2-2\sqrt{(2x-4)(x+5)}=1$ using

${(a-b)}^2=a^2+b^2-2ab$

$\Rightarrow 2x-4+x+5-2\sqrt{(2x-4)(x+5)}=1$

$\Rightarrow 3x+1-2\sqrt{(2x-4)(x+5)}-1=0$

$\Rightarrow 3x-2\sqrt{(2x-4)(x+5)}=0$

$\Rightarrow 2\sqrt{(2x-4)(x+5)}=3x$

squaring both sides, we get

$4(2x-4)(x+5)=9x^2$

$\Rightarrow x^2-24x+80=0$ on simplifying

$\Rightarrow (x-20)(x-4)=0$

$\therefore x=20,4$

But for the value $x=20,$ we have

$\sqrt{2x-4}-\sqrt{x+5}=\sqrt{2\times 20-4}-\sqrt{20+5}=\sqrt{40-4}-\sqrt{25}=\sqrt{16}-\sqrt{25}=4-5\ne 1$

and for $x=4$ we have

$\sqrt{2x-4}-\sqrt{x+5}=\sqrt{2\times 4-4}-\sqrt{4+5}=\sqrt{8-4}-\sqrt{9}=\sqrt{4}-\sqrt{9}=2-3\ne 1$

Since L.H.S$\ne$ R.H.S $x=4,20$ is not the solution to above problem.