Question

Solve the following equations:
(i) $\sqrt{2x-4}-\sqrt{x+5}=1$
(ii)$\sqrt{x}+\sqrt{x-\sqrt{1-x}}=1$
(iii)$x^2-4x+\left[x\right]+3=0$

where $\left[x\right]$ denotes the integral part of $x$

Answer 1

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

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

$\left(ii\right)\sqrt{x}+\sqrt{x+\sqrt{1-x}}=1$

$\Rightarrow \sqrt{x+\sqrt{1-x}}=1-\sqrt{x}$

Squaring both sides, we get

$\Rightarrow x+\sqrt{1-x}=1+x-2\sqrt{x}$

$\Rightarrow \sqrt{1-x}=1-2\sqrt{x}$

Squaring on both sides, we get

$\Rightarrow 1-x=1+4x-4\sqrt{x}$

$\Rightarrow -5x=-4\sqrt{x}$

Squaring both sides, we get

$\Rightarrow 25x^2=16x$

$\Rightarrow x(25x-16)=0$

$\therefore x=0,\frac{16}{25}$

For $x=0,\sqrt{x}+\sqrt{x+\sqrt{1-x}}=0+\sqrt{0+\sqrt{1-0}}=1$

For $x=\frac{16}{25},\sqrt{\frac{16}{25}}+\sqrt{\frac{16}{25}+\sqrt{1-\frac{16}{25}}}$

$=\frac{4}{5}+\sqrt{\frac{16}{25}+\sqrt{\frac{25-16}{25}}}$

$=\frac{4}{5}+\sqrt{\frac{16}{25}+\sqrt{\frac{9}{25}}}$

$=\frac{4}{5}+\sqrt{\frac{16}{25}+\frac{3}{5}}=\frac{4}{5}+\sqrt{\frac{16+15}{25}}=\frac{4}{5}+\frac{31}{25}\ne 1$

Hence $x=0$ is the only solution.

$\left(iii\right)x^2-4x+1+3=0$

$\Rightarrow x^2-4x+4=0$

$\Rightarrow x^2-2\times x\times 2+2^2=0$

$\Rightarrow {(x-2)}^2=0$

$\therefore x=2$