Chapter 4.7, Problem 3CFU

To determine

The difference between solving an equation with one radical and solving an equation with more than one radical equations.

Explanation of Solution

An equation can have one or more than one radical. If an equation has one radical equation, then the solution is done by isolating the radical first. It is because when we isolate the radical, then the radical comes in one side of the equation. Then by squaring both sides of the equation, we can remove the square root of the radial. After removing the square root we can simplify the equation and then solve it.

If an equation has more than one radical equation, then the solution is done by taking one radical in each side of the equation first. Then by squaring both sides of the equation, we can remove the square root of the radial. After removing the square root we can simplify the equation and then solve it.

Now we take an example of one radical 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 take an example of more than one radical equation,

  $\begin{array}{l}\sqrt{x+1}-\sqrt{2x+3}=0\\ \Rightarrow \sqrt{x+1}=\sqrt{2x+3}\\ \Rightarrow {\left(\sqrt{x+1}\right)}^2={\left(\sqrt{2x+3}\right)}^2\\ \Rightarrow x+1=2x+3\\ \Rightarrow 2x-x=1-3\\ \Rightarrow x=-2\end{array}$