Question

solve the radical equation. check your solution(s). write
o solution\ if there is no solution. (sqrt{26 - 2x} - sqrt{2x - 10} = 0) (x = square)

Explanation:

Step1: Isolate one radical

Given the equation $\sqrt{26 - 2x} - \sqrt{2x - 10} = 0$, we can add $\sqrt{2x - 10}$ to both sides to get $\sqrt{26 - 2x} = \sqrt{2x - 10}$.

Step2: Square both sides

Squaring both sides of the equation $\sqrt{26 - 2x} = \sqrt{2x - 10}$ gives us $26 - 2x = 2x - 10$.

Step3: Solve for x

First, add $2x$ to both sides: $26 = 4x - 10$. Then, add $10$ to both sides: $36 = 4x$. Divide both sides by $4$: $x = 9$.

Step4: Check the solution

We need to check if $x = 9$ makes the expressions under the square roots non - negative.
For $\sqrt{26 - 2x}$: When $x = 9$, $26-2\times9=26 - 18 = 8>0$.
For $\sqrt{2x - 10}$: When $x = 9$, $2\times9-10 = 18 - 10 = 8>0$.
Also, substitute $x = 9$ into the original equation:
Left - hand side: $\sqrt{26-2\times9}-\sqrt{2\times9 - 10}=\sqrt{8}-\sqrt{8}=0$, which is equal to the right - hand side.

Answer:

$9$