Ab = (7x-6) → b = (7x-6)/a (1)
bc = (12-2x) → b = (12-2x)/c (2)
Since (1) = (2) → (7x-6)/a = (12-2x)/c OR a/c = (7x-6)/(12-2x) (3)
Multiply both numerators of (3) by the SQUARE of their respective denominators;
(a*c²)/c = (7x-6)(12-2x)²/(12-2x)
Now simplify:
ac = (7x-6)(12x-2x) or ac = -14x² + 96x - 72
The property that illustrates (ab)3 = a(b3) is
ASSOCIATIVE PROPERTY OF MULTIPLICATION
Answer:
no real solution is the answer
If you have to list your extraneous solutions somewhere, that would be x=0.
Step-by-step explanation:
Cool thing here is the radical, the thing with the square root, is already isolate.
So we need to square both sides as a first step. (Whenever you raise both sides to even power, you must definitely check your solutions as some might not actually be solutions to the original equation)
So upon squaring both sides I get:
x^4+16=(x^2-4)^2
Now I will write (x^2-4)^2 as (x^2-4)(x^2-4) and foil it! This gives me:
x^4+16=x^4-8x^2+16
Subtract x^4 and 16 on both sides:
0=-8x^2
Divide both sides by -8
0=x^2
Therefore x=0.
0 isn't actually a solution though because when you plug it in you get 4=-4 which is not true.
