Sign In
Do not forget to check for extraneous solutions.
Consider both sides of the equation as functions. To graph the left-hand side of the equation, begin by determining the transformations.
Is there any extraneous solution when you solved the equation graphically?
x=3
x=3
See solution.
to solve the equation algebraically, we will first isolate the radical term. Then, we can continue by squaring the equation write the resulting equation standard form.
LHS-3=RHS-3
LHS^2=RHS^2
(a-b)^2=a^2-2ab+b^2
LHS+3=RHS+3
LHS-x=RHS-x
Rearrange equation
Next, we will factor the equation by grouping.
Finally, we can solve it by the Zero Product Property.
Use the Zero Product Property
(I): LHS+4=RHS+4
(II): LHS+3=RHS+3
The solutions to the equation are x=3 and x=4. However, since we have a radical in the equation, we should check for the extraneous solutions.
| x | 3-sqrt((x-3))? =x | Result |
|---|---|---|
| 3 | 3-sqrt(( 3-3))? = 3 | 3=3 |
| 4 | 3-sqrt(( 4-3))? = 4 | 2≠4 |
Because x=4 does not satisfy the equation, the only solution is x=3.
To solve the equation graphically, we will consider both sides of the equation as functions.
3-sqrt((x-3))=x
⇓
f(x)=g(x)
With this, we can determine the functions as f(x)=3-sqrt((x-3)) and g(x)=x. Because g(x)=x is an identity function, we can graph it immediately.
To graph f(x), we should first determine the transformations that is applied to the parent function y=sqrt(x) to get f(x). Let's consider the possible transformations.
| Transformations of y=sqrt(x) | |
|---|---|
| Vertical Translations | Translation up k units, k>0 y=sqrt(x)+ k |
| Translation down k units, k>0 y=sqrt(x)- k | |
| Horizontal Translations | Translation right h units, h>0 y=sqrt(x- h) |
| Translation left h units, h>0 y=sqrt(x+ h) | |
| Reflections | In the x-axis y=- sqrt(x) |
| In the y-axis y=sqrt(- x) | |
Using the table, the transformations that are applied to the parent function to get f(x)=-sqrt((x-3))+3 can be listed as below.
Now, we can graph f(x) and g(x) on the same coordinate plane. The point(s) of intersection will determine the solution(s) of the equation.
Therefore, the only solution to the equation is x=3.
When we solved the equation algebraically, we found two solutions. However, one of them was an extraneous solution. On the other hand, when we solved it graphically, we only found one solution and there was no extraneous solution.