Chapter 7, Problem 16PT

To determine

To Solve:

  $\sqrt{x-5}=\sqrt{2x-4}$

Answer to Problem 16PT

  $No\text{ solution}$

Explanation of Solution

Given Information:

The given equation is :

  $\sqrt{x-5}=\sqrt{2x-4}$

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.
• Definition of nth root:
• Words : For any real numbers a and b , and any positive integer n, if $a^n=b$ then a is the nth root of b. That is $a=\sqrt[n]{b}$
• $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$
• The non-negative root is called the Principal Root.
• Suppose n is an integer greater than 1, and a is a real number.

a	n is even	n is odd
a >0	1 unique positive and 1 unique negative real root : $\pm \sqrt[n]{a}$ ; positive root is principal root.	1 unique positive and 0 negative real root : $\sqrt[n]{a}$
a < 0	0 real roots	0 positive and 1 negative real root : $-\sqrt[n]{a}$
a=0	1 real root : $\sqrt[n]{0}=0$	1 real root : $\sqrt[n]{0}=0$

Calculation:

  $\begin{array}{l}Given\\ \sqrt{x-5}=\sqrt{2x-4}\\ \Rightarrow {\left(\sqrt{x-5}\right)}^2={\left(\sqrt{2x-4}\right)}^2\mathrm{.......}[\text{eliminate the radical ]}\\ \Rightarrow x-5=2x-4\\ \Rightarrow x-5+4-x=2x-4+4-x\mathrm{............}[\text{add (4-x) each side]}\\ \Rightarrow -1=x\\ \Rightarrow x=-1\\ Verify:\\ \text{Substitute x = -1 in the given equation:}\\ \sqrt{x-5}=\sqrt{2x-4}\\ \sqrt{-1-5}=\sqrt{2(-1)-4}\\ \sqrt{-6}=\sqrt{-6}\\ But\text{ the square root is not defined for a negative radicand}\text{.}\\ \text{Here the radicand is -6 , so this is not defined}\text{.}\\ So,\text{ x=-1 is not the solution of the given equation}\text{.}\end{array}$