Functions: Power functions and root functions
Root equations
Here are examples where a root equation needs to be solved.
#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#.
\[\begin{array}{rcl}\sqrt{4x+5}&=&3 \\ &&\phantom{xxx}\blue{\text{the original equation}}\\
4x+5&=&9 \\ &&\phantom{xxx}\blue{\text{both sides squared}}\\
4x&=& 4 \\ &&\phantom{xxx}\blue{\text{both sides minus }5}\\
x&=&1 \\ &&\phantom{xxx}\blue{\text{both sides divided by }4}\end{array}\]
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.
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. 
# x={{7}\over{8}} #
\[\begin{array}{rcl}
2\cdot \sqrt{x}&=& \sqrt{74\cdot x}\\
&&\phantom{xxx}\blue{\text{original equation}}\\
4\cdot x&=&74\cdot x \\
&&\phantom{xxx}\blue{\text{both sides squared}} \\
8\cdot x&=&7 \\
&&\phantom{xxx}\blue{\text{terms of }x \text{ brought to the left, constant terms brought to the right}} \\
x&=&{{7}\over{8}} \\
&&\phantom{xxx}\blue{\text{divided by the coefficient of }x} \\
\end{array}\]
\[2\cdot \sqrt{{{7}\over{8}}}={{\sqrt{7}}\over{\sqrt{2}}}\]
On the right side is:
\[\sqrt{74\cdot \left({{7}\over{8}}\right)}={{\sqrt{7}}\over{\sqrt{2}}}\]
Left and right are equal, so this solution is correct.
In conclusion, the answer of the equation is # x={{7}\over{8}} #.
\[\begin{array}{rcl}
2\cdot \sqrt{x}&=& \sqrt{74\cdot x}\\
&&\phantom{xxx}\blue{\text{original equation}}\\
4\cdot x&=&74\cdot x \\
&&\phantom{xxx}\blue{\text{both sides squared}} \\
8\cdot x&=&7 \\
&&\phantom{xxx}\blue{\text{terms of }x \text{ brought to the left, constant terms brought to the right}} \\
x&=&{{7}\over{8}} \\
&&\phantom{xxx}\blue{\text{divided by the coefficient of }x} \\
\end{array}\]
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:\[2\cdot \sqrt{{{7}\over{8}}}={{\sqrt{7}}\over{\sqrt{2}}}\]
On the right side is:
\[\sqrt{74\cdot \left({{7}\over{8}}\right)}={{\sqrt{7}}\over{\sqrt{2}}}\]
Left and right are equal, so this solution is correct.
In conclusion, the answer of the equation is # x={{7}\over{8}} #.
Unlock full access
Teacher access
Request a demo account. We will help you get started with our digital learning environment.
Student access
Is your university not a partner?
Get access to our courses via Pass Your Math independent of your university. See pricing and more.
Or visit omptest.org if jou are taking an OMPT exam.
Or visit omptest.org if jou are taking an OMPT exam.