How do you solve #x= sqrt(6x) +7#?
2 Answers
Explanation:
First let's define the domain :
So:
So :
\0/ herex's our answer!
Bare with me please: this is a long solution process:
Explanation:
We can subtract
We can square both sides to get:
What I have in blue, we can use the highly useful mnemonic FOIL to simplify this. We simply multiply the first terms, outside terms, inside and last terms. We get:
- First terms:
#x*x=x^2# - Outside terms:
#x*-7=-7x# - Inside terms:
#-7*x=-7x# - Last terms:
#-7*-7=49#
Now, we have:
Which can be simplified to
We have a quadratic on the left, so we want to set it equal to zero to find its zeroes. We do this by subtracting
The only factors of
where
Plugging into the quadratic formula, we get:
Which simplifies to
Because
Which obviously simplifies to
Now, we have
Every term is divisible by
Which is equal to
Since our answer had a square root in it, we know the domain has to be
as our final solution.
This was a long solution process, but all I did was:
- Get
#x# on one side - Square both sides to get rid of the square root
- Used FOIL to simplify the left side
- Used the Quadratic Formula
- Checked the domain
I really hope this helps!