Solving More Complex Square Root Inequalities
Description/Explanation/Highlights
Video Description
This video explains how to solve more complex square root inequalities that will result in multiple domain restrictions and squaring the original equation more than once.
Steps and Key Points to Remember
To solve a more complex square root inequality, follow these steps:
- Square root inequalities are solved in the same way as equations except when multiplying or dividing by a negative number as part of finding the solution, the inequality sign must be flipped. If you need help with solving basic square root inequalities, please refer to the video and page, Solving Basic Square Root Inequalities.
- In addition, when solving a square root inequality, we must first take into account limits that may exist to the domain that would cause a number under the square root to be negative.
- To find limits to the domain, set the portion of the equation under the square root sign \(\geq 0\) and solve the inequality. If there is more than one part under square root signs, set up each as a separate inequality. Any answer that you get from solving the entire inequality may be limited by these domain restrictions.
- To solve a square root inequality:
- Isolate the square root.
- Square both sides to get rid of the square root.
- If there is still a square root sign after squaring, repeat the process of isolating and squaring.
- Solve the resulting inequality remembering to flip the inequality sign if division or multiplication by a negative number is required.
Compare the solution set to the limits to the domain you found above to see if the solution must be limited.
- To solve the square root inequality \(\sqrt{x}+\sqrt{x-5}\leq 5\), first find limits to the domain by setting both parts under the square root signs \(\geq0\), \(x\geq 0\) and \(x-5\geq 0\).
- This will result in domain restrictions of \(x\geq0\) AND \(x\geq5\). Since both must be true, only numbers greater than or equal to 5 will make both true so the domain restriction is \(x\geq5\).
- Solve the original equation by isolating one of the square roots \(\sqrt{x-5}\leq-\sqrt{x}+5\) and squaring both sides \(\sqrt{x-5}^2\leq(-\sqrt{x}+5)^2\).
- The result will still leave a square root in the equation so the process must be repeated. \(x-5\leq x-10\sqrt{x}+25\longrightarrow 3\geq\sqrt{x}\). Note how the sign flipped because we had to divide by a negative number (-10) to isolate the square root.
- Solving this inequality gives us \(x\leq 9\). Since the domain restriction is \(x\geq5\), the only numbers in the solution set that allowed are those between 5 and 9, including 5 and 9.
- The final solution is written as \(5\leq x\leq9\) or in interval notation, \([5,9]\).
- The square root equation, \(\sqrt{x-1}+3>x\) will result in a quadratic after isolating and squaring. These are handled differently.
- Like any square root equation, determine the domain restrictions first. In this example after setting \(x-1\geq 0\), the domain is \(x\geq 1\).
- The quadratic that results from squaring and simplifying is set up as an equation equal to zero for the purpose of finding “boundaries.”
- In this case, the resulting quadratic is \(x^2-7x+10=0\) which factors to \((x-5)(x-2)=0\) so the boundary points are \(x=5\) and \(x=2\) but when plugging in to check, we find that 2 is an extraneous solution.
- 5 is a boundary as is the domain of \(x\geq1\). Try test points to see if the solution set is between 1 and 5 or greater than 5. Plugging 2 into the original gets a true statement but plugging 10 in does not so the solution set is between 1 and 5 but does not include 5 since the original problem was only greater than.
- The solution for this example is \(1\leq x<5\) or in interval notation, \([1,5)\).
Here are some key points to keep in mind when solving more complex square root inequalities.
- Square root inequalities may have more than on square root. If so, each part under square root signs should be set greater than or equal to zero to determine multiple domain restrictions.
- All domain restrictions must be considered together to determine the final domain that includes all solutions.
- Square root inequalities that result in quadratics after squaring and simplifying, are handled differently. The quadratic is set up as an equation to find “boundaries” to test.
- Boundaries are first checked to see if they are extraneous and then boundaries are combined with the domain and test points applied to find the solution set.
- It may be helpful to watch the video to see the entire process.
Video Highlights
- 00:00 Introduction
- 00:33 \(\sqrt{x}+\sqrt{x-5}\leq 5\) example of solving a square root inequality with multiple domain restrictions
- 06:38 \(\sqrt{x-1}+3>x\) example of solving a square root inequality that results in a quadratic after isolating and squaring
- 13:42 Conclusion
- To watch this video on YouTube in a new window with clickable highlights, click here