Root equations

Functions: Power functions and root functions

Root equations

Here are examples where a root equation needs to be solved.

Solve #x# in
\[\sqrt{4x+5}=3\]
Give your answer as #x=x_1 \lor x=x_2 \lor \ldots \lor x=x_n#, in which #x_1#, #x_2#, #\ldots# #x_n# are numbers and #n# the number of solutions.

#x=1#

\[\begin{array}{rcl}\sqrt{4x+5}&=&3 \\ &&\phantom{xxx}\blue{\text{the original equation}}\\
4x+5&=&9 \\ &&\phantom{xxx}\blue{\text{both sides squared}}\\
4x&=& 4 \\ &&\phantom{xxx}\blue{\text{both sides minus }5}\\
x&=&1 \\ &&\phantom{xxx}\blue{\text{both sides divided by }4}\end{array}\]

Now we check the found solutions #x=1# by entering it into the equation:
\[\sqrt{4 \cdot 1+5}=\sqrt{9}=3\]
Hence, the solution is correct and the solution to the equation is #x=1#.

In the coordinate system below, we see the graph #f(x)=\sqrt{4x+5}# in blue (solid) and #g(x)=3# in green (dashed), and their intersection point #\rv{1,3}# in red.

New example

In general we can solve a root equation with the following #4# steps.

Solving root equations

	Procedure We solve a root equation for #x#.	Example #\sqrt{x+4}+4=9#
Step 1	Isolate the root. This means that by means of reduction we make sure the root is the only thing on one side of the equation.	#\sqrt{x+4}=5#
Step 2	Take the square on both sides to get rid of the root.	#x+4=25#
Step 3	Solve this equation.	#x=21#
Step 4	Check if the found solution is a solution to the original equation.	#\sqrt{21+4}+4=9# Hence, the solution is correct.

CheckingChecking the solution(s) is necessary, because squaring something can result in solutions that are not solutions to the original equation. This can be seen in the second example on top of this page.

Solve the equation below:
\[\sqrt{5\cdot x-2}=\sqrt{8-x}\]
Give your answer in the form of

#none# if there is no solution,
#x=x_1# if there is one solution,
#x=x_1\lor x=x_2# if there are two solutions,

with the correct values for #x_1# and #x_2#.

# x={{5}\over{3}} #

\[\begin{array}{rcl}
\sqrt{5\cdot x-2}&=& \sqrt{8-x}\\
&&\phantom{xxx}\blue{\text{original equation}}\\

5\cdot x-2&=&8-x \\
&&\phantom{xxx}\blue{\text{both sides squared}} \\
6\cdot x&=&10 \\
&&\phantom{xxx}\blue{\text{terms of }x \text{ brought to the left, constant terms brought to the right}} \\

x&=&{{5}\over{3}} \\
&&\phantom{xxx}\blue{\text{divided by the coefficient of }x} \\
\end{array}\]

Because we have taken the square, the solution for #x# that we found may not be a solution to the original equation. Therefore, we must now test the solution we found by entering it into the original equation.

On the left side is:
\[\sqrt{5\cdot \left({{5}\over{3}}\right)-2}={{\sqrt{19}}\over{\sqrt{3}}}\]
On the right side is:
\[\sqrt{8-{{5}\over{3}}}={{\sqrt{19}}\over{\sqrt{3}}}\]
Left and right are equal, so this solution is correct.

In conclusion, the answer of the equation is # x={{5}\over{3}} #.

New example