## Related questions with answers

Solve the equation.

$\sqrt { x + 6 } - \sqrt { 2 x - 4 } = 0$

Solution

VerifiedThis is radical equation with two radicals. First you should add $\sqrt{2x-4}$ to the both sides of given equation:

$\begin{align*} \sqrt{x+6}-\sqrt{2x-4}+\sqrt{2x-4}&=0+\sqrt{2x-4}\\ \sqrt{x+6}&=\sqrt{2x-4}\\ (\sqrt{x+6})^2&=(\sqrt{2x-4})^2\\ x+6&=2x-4\tag{$(\sqrt{a})^2=a$, for $a\geq 0$}\\ x+6-2x&=2x-4-2x\tag{Substitute $2x$}\\ -x+6&=-4\tag{Simplify}\\ -x+6-6&=-4-6\tag{Subtract $6$}\\ -x&=-10\tag{Simplify}\\ x&={\color{#4257b2}10}\tag{Multiply by $-1$} \end{align*}$

Check (substitute solution in given equation):

$\begin{align*} \sqrt{ {\color{#4257b2}10}+6}&=\sqrt{ 2\cdot {\color{#4257b2}10}-4}\\ \sqrt{16}&=\sqrt{20-4} \\ \sqrt{16}&=\sqrt{16} \\ 4&=4 \quad {\color{#c34632}\checkmark} \\ \end{align*}$

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