Solving Square Root Equations
Description/Explanation/Highlights
Video Description
This video explains how to solve a square root equation.
Steps and Key Points to Remember
To solve a square root equation, follow these steps:
- Square root equations are equations that have a variable under the square root sign. \(3\sqrt{x-4}+2=32\) is a square root equation because x is under the square root symbol.
- To solve a square root equation:
- Isolate the square root.
- Square both sides to get rid of the square root.
- Solve the resulting equation.
If the equation is non-linear, check for extraneous solutions.
- To solve the equation \(3\sqrt{x-4}+2=32\), first isolate the square root by subtracting 2 from both sides and dividing by 3.
- That will leave \(\sqrt{x-4}=10\) with the square root isolated.
- Now square both sides. This will get rid of the square root sign on the left and square 10 leaving, \(x-4=100\).
- Solve the simple linear equation and the solution is \(x=104\).
- Although checking is always a good idea, it is not necessary in this equation since the result of squaring both sides was a linear equation.
- The equation \(\sqrt{x+5}-3=x\) is different, however, since adding 3 to both sides and squaring both sides results in a quadratic equation.
\(\sqrt{x+5}-3=x\)
\(\sqrt{x+5}=x+3\) <==result of adding 3 to both sides
\({\sqrt{x+5}}^2=(x+3)^2\) <==square both sides [remember to square a binomial, expand it to \((x+3)(x+3)\)]
\(x+5=x^2+6x+9\) <==the results of squaring both sides is a quadratic equation since there is an \(x^2\).
\(0=x^2+5x+4\) <==get all the variables and numbers to one side and set equal to 0.
\(0=(x+4)(x+1)\) <==factor the quadratic
\(x+4=0\) & \(x+1=0\) <==set each factor equal to 0 and solve for x.
\(x=-4\) & \(x=-1\) <==possible solutions for the square root equation
- Since squaring both sides resulted in a quadratic equation (non-linear), we must check for extraneous solutions.
- Plug in -4 for x in the original equation. The result will be -2 = -4 which is a false statement. \(x=-4\) is therefore an extraneous solution and must therefore be eliminated. Plugging in -1 for x however gets a result of -1 = -1 which is a true statement and therefore \(x=-1\) is a solution (and in fact the only solution).
- Sometimes you may need to square both sides more than once to eliminate all of the square roots. \(\sqrt{x+2}+1=\sqrt{3-x}\) is an example of an equation that will require squaring more than once.
- Notice that the square root is already isolated on the right side of the equation so we should square both sides. \((\sqrt{x+2}+1)^2=(\sqrt{3-x})^2\).
- This eliminates the square root on the right but the left side must be written as two binomials and multiplied together. \((\sqrt{x+2}+1)(\sqrt{x+2}+1)=3-x\longrightarrow x+2+2\sqrt{x+2}+1=3-x\).
- Now isolate the square root on the left side by combining like terms, and getting everything not under the square root sign to the right side by subtracting and dividing leaving \(\sqrt{x+2}=-x\).
- The square root on the left side is now isolated so we can square both sides once again \((\sqrt{x+2})^2=(-x)^2\) which gives us \(x+2=x^2\), a quadratic.
- Get everything on one side and factor. \(x^2-x-2=0\longrightarrow (x-2)(x+1)=0\).
- \(x=2\) & \(x=-1\) are possible solutions but when 2 is substituted into the original equation, a false statement is created when simplified so only \(x=-1\) is a valid answer and \(x=2\) is an extraneous solution.
Here are some key points to keep in mind when solving square root equations.
- Square root equations are recognized by a variable or variables under a square root sign or signs on one or both sides of the equation.
- To solve: Isolate a square root on one side of the equation, square both sides to eliminate the square root, solve the resulting equation, and check for extraneous solutions.
- If squaring both sides does not eliminate all the square roots, the process of isolating and squaring may have to be repeated.
- Extraneous solutions are solutions that are not valid. To check for extraneous solutions, plug each potential solution into the original equation and simplify to see if the resulting statement is true. If it is not true, the solution is extraneous and must be eliminated.
- If the equation after squaring both sides is linear, extraneous solutions will not occur and checking is not required.
Video Highlights
- 00:00 Introduction
- 00:05 Steps for solving a square root equation
- 01:30 \(3\sqrt{x-4}+2=32\) example of solving a square root equation
- 03:55 \(\sqrt{x+5}-3=x\) example of solving a square root equation with an extraneous solution
- 08:57 \(\sqrt{x+2}+1=\sqrt{3-x}\) example of solving a square root equation that requires squaring more than once
- 14:44 Conclusion
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