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Let's start by replacing the variable b with the variable x. 2b=sqrt(b+3) b= x ⟶ 2 x=sqrt(x+3) Since the radical is isolated, we can raise both sides of the equation to a power equal to the index of the radical. Let's try to solve our equation using this method!
LHS^2=RHS^2
(a b)^m=a^m b^m
Calculate power
( sqrt(a) )^2 = a
Substitute values
Using the Quadratic Formula, we found that the solutions of the given equation are x= 1±78.
| x=1± 7/8 | |
|---|---|
| x_1=1+7/8 | x_2=1-7/8 |
| x_1=8/8 | x_2=-6/8 |
| x_1= 1 | x_2= -3/4 |
Therefore, the solutions are x_1= 1 and x_2= - 34. Recall that, at the beginning of the solution, we substituted x for b. Consequently, the solutions to the original equation are b_1= 1 and b_2= - 34. Let's check them to see if we have any extraneous solutions.
We will check b_1=1 and b_2=- 34 one at a time.
Let's substitute b= 1 into the original equation.
b= 1
Identity Property of Multiplication
Add terms
Calculate root
We got a true statement. Therefore, b=1 is a solution of the original equation.
Now, let's substitute b= - 34.
b= - 34
Put minus sign in numerator
a*b/c= a* b/c
a(- b)=- a * b
Multiply
Put minus sign in front of fraction
a/b=.a /2./.b /2.
a = 4* a/4
Commutative Property of Addition
Subtract fractions
Subtract terms
Calculate root
In this case, we got a false statement, so b=-3/4 is an extraneous solution. Therefore, b=1 is the only solution of the original equation.