We will explain how to an that contains . To do so, we will consider the variety of places where expressions, , and can be found in equations.
Radical on One Side
We will classify the equations that have a radical on one side depending on the existence of variables outside the radical. No matter if there are variables outside the radical or not, we will solve both types of the equations by applying the following steps.
- Isolate the radical on one side of the equation.
- Square both sides of the equation.
- Solve the equation for the variable.
- Substitute the into the original equation to check whether it is or not.
Without Variables Outside the Radical
Let's first consider the situation that the only variable is under the radical sign. We will write an example for this case and solve it to see the solution process.
3 sqrt(x+2)+4=19
Let's start by isolating the radical on the left side of the equation. To do so, we can use the .
3 sqrt(x+2)+4=19
3 sqrt(x+2)=15
sqrt(x+2)=5
Next, we will square both sides of the equation to get rid of the square root.
( sqrt(x+2))^2=5^2
x+2 =25
Then, we can solve the equation to find x.
At last, we need to substitute the solution back into the original equation to check whether it is extraneous or not.
3sqrt(x+2)+4=19
3sqrt(23+2)+4 ? = 19
3sqrt(25)+4 ? = 19
3(5) +4 ? = 19
15+4 ? = 19
19 = 19 âś“
With Variables Outside the Radical
Next, we will examine radical equations that include some variables outside the radical. Let's write an example!
x+ sqrt(- x +14)=2
One more time, we will begin by isolating the radical on one side of the equation. To do so, let's subtract x from both sides of the equation.
x+ sqrt(- x +14)=2 ⇒ sqrt(- x +14)=2- x
Then, we can square both sides of the equation to get rid of the radical sign.
( sqrt(- x +14))^2=(2- x)^2
- x+14=(2- x)^2
- x+14=4-4 x+ x^2
4-4 x+ x^2=- x+14
4-3 x+ x^2=14
-10-3 x+ x^2=0
x^2-3 x-10=0
x^2-3 x-10=0
x=- ( - 3)±sqrt(( - 3)^2-4( 1)( -10))/2( 1)
x=3±sqrt(( - 3)^2-4( 1)( -10))/2( 1)
x=3±sqrt(9+ 40)/2
x=3±sqrt(49)/2
x=3± 7/2
lc x= 3+72 & (I) x= 3 - 72 & (II)
(I), (II):Add and subtract terms
l x= 102 x= -42
(I), (II):Calculate quotient
lx_1=5 x_2=-2
|
|
Substitution
|
Check
|
| x_1=5
|
5+ sqrt(-( 5 )+14) ? =2
|
8 ≠2 *
|
| x_2=-2
|
-2+ sqrt(-( -2 )+14) ? = 2
|
2=2 âś“
|
Notice that although we found two solutions for the equation, 5 does not satisfy the original equation. Therefore, 5 is an extraneous solution and -2 is a real solution.
Radicals on Both Sides
This time, we will examine radical equations that include radicals on both sides of the equation. The solution process for equations that include only radicals will be different from equations that also include constants, we can write some general steps to solve both types of the equations.
- Square both sides of the equation until the all radical signs are eliminated.
- Solve the equation for the variable.
- Substitute the solution into the original equation to check whether it is extraneous or not.
Only Radicals
We will begin by writing an example involving only square roots on both sides of the equation.
sqrt(2 x-3) = sqrt(x-1)
Since the radicals are already isolated in different sides of the equation, we will square both sides of the equation to get rid of the radical signs.
( sqrt(2 x-3))^2=( sqrt(x-1))^2
2 x-3= x-1
sqrt(2x-3)=sqrt(x-1)
sqrt(2( 2)-3) ? = sqrt(2-1)
sqrt(4-3) ? = sqrt(2-1)
sqrt(1) ? = sqrt(1)
1 = 1 âś“
Radicals and Constants
As our last case, we will examine equations that include constants along with the radicals on both sides of the equation.
2- sqrt(x-3)= sqrt(x)
We can start by squaring both sides of the equation.
(2- sqrt(x-3))^() 2=( sqrt(x))^() 2
4-4 sqrt(x-3)+ x-3=( sqrt(x))^() 2
4-4 sqrt(x-3)+ x-3= x
4-4 sqrt(x-3)-3=0
4-3=4 sqrt(x-3)
1=4 sqrt(x-3)
1/4= sqrt(x-3)
(1/4)^2=( sqrt(x-3) )^() 2
1/16= x-3
1/16+3= x
1/16+48/16= x
49/16= x
x=49/16
2-sqrt(x-3)=sqrt(x)
2-sqrt(49/16-3) ? =sqrt(49/16)
2-sqrt(49/16-48/16) ? =sqrt(49/16)
2-sqrt(1/16) ? =sqrt(49/16)
2-1/4? = 7/4
8/4-1/4? = 7/4
7/4 = 7/4 âś“
Conclusion
To summarize the steps for solving an equation that contains square roots, let's review how we solved each of the example equations.
