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To solve equations with a variable expression inside a radical, we first want to make sure the radical is isolated. Then 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
( sqrt(a) )^2 = a
Rearrange equation
We now have a quadratic equation, and we need to find its roots. To do it, let's identify the values of a, b, and c. d^2-d-12=0 ⇕ 1d^2+( - 1)d+( -12)=0
Substitute values
- (- a)=a
(- a)^2=a^2
Identity Property of Multiplication
- a(- b)=a* b
Add terms
Calculate root
Using the Quadratic Formula, we found that the solutions of the given equation are d= 1± 7 2.
| d=1± 7/2 | |
|---|---|
| d_1=1+7/2 | d_2=1-7/2 |
| d_1=8/2 | d_2=-6/2 |
| d_1= 4 | d_2= -3 |
Therefore, the solutions are d_1=4 and d_2=-3. Let's check them to see if we have any extraneous solutions.
We will check d_1=4 and d_2=-3 one at a time.
Let's substitute d=4 into the original equation.
In this case we got a true statement. Therefore, d=4 is a solution of the original equation.
Now, let's substitute d= -3.
We got a false statement, so d=-3 is an extraneous solution. Therefore, d=4 is the only solution of the original equation.