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LHS-2=RHS-2
LHS^2=RHS^2
( sqrt(a) )^2 = a
(a-b)^2=a^2-2ab+b^2
Calculate power and product
Let's now apply the Zero Product Property to find the solutions to the equation.
Use the Zero Product Property
(I): LHS+1=RHS+1
(II): LHS+4=RHS+4
We found that the solutions to the equation are 1 and 4. Finally, we will check our solutions by substituting the value into the original equation. Let's start with x=1.
Since substituting x=1 created a false statement, x=1 is not a solution to the given equation. Let's now check for x=4.
This time we created a true statement. Therefore, x=4 is indeed a solution to the given equation.
LHS^2=RHS^2
(a+b)^2=a^2+2ab+b^2
( sqrt(a) )^2 = a
Calculate power and product
We found that the solution to the equation is x= 14. Finally, we will check our solution by substituting x= 14 into the original equation.
x= 1/4
a = 4* a/4
Add fractions
sqrt(a/b)=sqrt(a)/sqrt(b)
Calculate root
a = 2* a/2
Add fractions
Since substituting x= 14 created a true statement, x= 14 is indeed a solution to the given equation.