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Question

# Solve the equation.$\sqrt { x + 6 } - \sqrt { 2 x - 4 } = 0$

Solution

Verified
Step 1
1 of 2

This 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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