Chapter 4.7, Problem 2CFU

To determine

The necessity to check the extraneous solutions in radical equations.

Explanation of Solution

An extraneous solutions is a solution that does not solve the original equation. In other words it can be said that an extraneous solutions does not give the actual result of the given equation. So it is always necessary to check or verify the result of the equation to determine whether the result is correct or extraneous.

Now let us check for extraneous solutions by solving a radical equation.

Now we solve the equation,

  $\begin{array}{l}5+\sqrt{x+1}=x\\ \Rightarrow \sqrt{x+1}=x-5\\ \Rightarrow {\left(\sqrt{x+1}\right)}^2={\left(x-5\right)}^2\\ \Rightarrow x+1=x^2-10x+25\\ \Rightarrow x^2-11x+24=0\\ \Rightarrow x^2-8x-3x+24=0\\ \Rightarrow x\left(x-8\right)-3\left(x-8\right)=0\\ \Rightarrow (x-3)(x-8)=0\\ Therefore,\\ either,\\ (x-8)=0\\ x=8.\\ or,\\ (x-3)=0\\ x=3.\end{array}$

Now we should verify the result of the equation to check for extraneous solution.

  $\begin{array}{l}5+\sqrt{x+1}=x\\ Taking,x=8.\\ \Rightarrow 5+\sqrt{8+1}=8\\ \Rightarrow 5+\sqrt{9}=8\\ \Rightarrow 5+3=8\\ \Rightarrow 8=8\\ \text{Therefore we can see that LHS=RHS}\text{. Therefore the result is correct}\text{.}\\ 5+\sqrt{x+1}=x\\ Taking,x=3.\\ LHS\\ \Rightarrow 5+\sqrt{3+1}\\ \Rightarrow 5+\sqrt{4}\\ \Rightarrow 5+2\\ \Rightarrow 7\\ \text{Therefore we can see that LHS}\ne \text{RHS}\text{. Therefore the result is not correct}\text{.}\\ \text{Therefore, }x=3\text{ is an extraneous solution}\text{. }\end{array}$