Question

$\sqrt{x+20}+\sqrt{x+4}=4\sqrt{x-1}$.

Answer 1

We are given, $\sqrt{x+20}+\sqrt{x+4}=4\sqrt{x-1}$

now, $\sqrt{x+20}=4\sqrt{x-1}-\sqrt{x+4}$

now, taking square on both side, we get

$(\sqrt{x+20}{)}^2=(4\sqrt{x-1}-\sqrt{x+4}{)}^2$

$\therefore x+20=16(\sqrt{x-1}{)}^2-2(4\sqrt{x-1}\left)\right(\sqrt{x+4})+(\sqrt{x+4}{)}^2$

$\therefore x+20=16(\sqrt{x-1}{)}^2-2(4\sqrt{x-1}\left)\right(\sqrt{x+4})+(\sqrt{x+4}{)}^2$

$\therefore x+20=16(x-1)-8\left(\sqrt{(x-1)(x+4)}\right)+(x+4)$

$\therefore x+20=16x-16-8\sqrt{(x-1)(x+4)}+x+4$

$\therefore 8\sqrt{(x-1)(x+4)}=16x-32$

Dividing whole equation by $8$, we get

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

now, again taking the square on both sides we get

$(x-1)(x+4)=(2x-4{)}^2$

$\therefore x^2+4x-x-4=4x^2-2\left(2x\right)\left(4\right)+(4{)}^2$

$\therefore x^2+3x-4=4x^2-16x+16$

$\therefore x^2+3x-4=4x^2-16x+16$

$\therefore 3x^2-19x+20=0$

$\therefore 3x^2-15x-4x+20=0$

$\therefore 3x(x-5)-4(x-5)=0$

$\therefore (3x-4)(x-5)=0$

so,

$x=4/3$ & $x=5$