Chapter 6, Problem 11PT

To determine

To Solve:

  $\sqrt{3x}=\sqrt{x-2}$

Answer to Problem 11PT

  $No\text{ solution}$

Explanation of Solution

Given Information:

The given equation is :

  $\sqrt{3x}=\sqrt{x-2}$

Concept Used :

Solve Radical equations:

• Isolate the radical on one side of the equation.
• Raise each side of the equation to a power equal to the index of the radical to eliminate the radical .
• Solve the resulting polynomial equation.Check your results.
• When solving radical equations, the result may be a number that does not satisfy the original equation . Such a number is called an extraneous solution.

Calculation:

  $\begin{array}{l}Given\\ \sqrt{3x}=\sqrt{x-2}\\ \text{⇒}{\left(\sqrt{3x}\right)}^2={\left(\sqrt{x-2}\right)}^2\mathrm{.......}[\text{eliminate the radical]}\\ \text{⇒}3x=x-2\\ \text{⇒}3x-x=x-2-x\mathrm{............}[\text{subtract 'x' from each side]}\\ \text{⇒}2x=-2\\ \text{⇒}\frac{2x}{2}=\frac{-2}{2}\mathrm{...............................}[\text{divide each side by 2]}\\ \text{⇒}x=-1\\ Verify:\\ substitute\text{ the value x = -1 in the original equation:}\\ \text{⇒}\sqrt{3x}=\sqrt{x-2}\\ \text{⇒}\sqrt{3(-1)}=\sqrt{-1-2}\\ \text{⇒}\sqrt{-3}=\sqrt{-3}\\ \text{But the square root cannot take negative radicand}\text{.}\\ \text{and here , the radicand is -3 , which is negative }\text{.}\\ \text{So, x = -1 is not the solution of the given equation}\text{.}\end{array}$