how would you solve 4(√(x^2-80))=2x?

Answer 1

first squre both sides:

16 (x^2-80)=4x^2

then multiply the 16 through:

16x^2 - 1280 = 4x^2

Now, collect like terms:

12x^2 = 1280

divide both sides by 12:

x^2 = 106.666

take the square root of both sides:

x=10.32796

Answer 2

solve 4(√(x^2-80))=2x

1st: find the domain of the equation

x^2 -80 ≥ 0
x^2 ≥ 80
Domain = ] - infinity , - √80] and [√80 , +infinity[
Now we want to get rid off the square root.
The square root of anything is ≥ 0
we want to square both sides to get rid off the square root, for this to be correct we nedd 2x to be ≥ 0 . domain is restricted to the overlap of the 2 domains

Domain = [√80 , +infinity[
16(x^2 -80) = 4x^2
12x^2 = 16*80
x^2 = 16*80/12 =106.67 = 10.328^2
x = 10.328

Answer 3

4(√(x^2-80))=2x, x > 0
Square both sides,
16(x^2-80) = 4x^2
Distribute after dividing both sides by 4,
4x^2-320 = x^2
Collect variables,
3x^2 = 320
x = √(320/3), since x > 0

Answer 4

16(x^2-80) = 4x^2
4(x^2- 80) = x^2
4x^2 - 320 = x^2
3x^2 = 320
x^2 = 320/3
x = +/-sqrt(320/3)