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Raise both sides of the radical equation to a power equal to the index of the radical.
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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
(a b)^m=a^m b^m
Calculate power
.LHS /2.=.RHS /2.
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. 2t^2-t-28=0 ⇕ 2t^2+( - 1)t+( -28)=0
Substitute values
- (- a)=a
(- a)^2=a^2
Multiply
- a(- b)=a* b
Add terms
Calculate root
Using the Quadratic Formula, we found that the solutions of the given equation are t= 1± 15 4.
| t=1± 15/4 | |
|---|---|
| t_1=1+15/4 | t_2=1-15/4 |
| t_1=16/4 | t_2=-14/4 |
| t_1= 4 | t_2= -7/2 |
Therefore, the solutions are t_1=4 and t_2=- 72. Let's check them to see if we have any extraneous solutions.
We will check t_1=4 and t_2=- 72 one at a time.
Let's substitute t=4 into the original equation.
In this case we got a true statement. Therefore, t=4 is a solution of the original equation.
Now, let's substitute t= - 72.
t= -7/2
a(- b)=- a * b
A * a/A= a
Add terms
Calculate root
We got a false statement, so t=- 72 is an extraneous solution. Therefore, t=4 is the only solution of the original equation.