Solve following equations.
y ^{-6} - 9 y ^{-3} + 8 = 0
Solution
y ^{-6} - 9 y ^{-3 }+ 8 = 0
For this part notice that,
-6 = 2 (-3)
and thus we do have an equation which is reducible to quadratic form. The substitution is,
u = y ^{-3 }u ^{2 } = ( y ^{-3} )^{2 } = y ^{-6}
The equation becomes,
u ^{2} - 9u + 8 = 0
(u - 8) (u -1) = 0 u = 1, u = 8
Now, y's is going to take a little more work here, however shouldn't be too bad.
u =1: ⇒ y ^{-3} = 1/ y^{3} =1 ⇒ y3 =1/1=1 ⇒ y = (1)^{1/3} = 1
u = 8: ⇒ y ^{-3 }= 1 /y^{3} = 8 ⇒ y^{3} = 1/8 ⇒ y = (1 /8)(1/3 ) =1/2
The two solutions to this equation are following
y= 1 and y = 1/2 .
In this case we acquire four solutions & two of them are complex solutions. Getting complex solutions out of these are in fact more common that this set of instance might recommend. The problem is that to obtain some of the complex solutions need knowledge which we haven't (and won't) cover in this course. Thus, they don't show up all that often.
All of the examples to this point gave quadratic equations that were factorable or in the case of the last part of the earlier example was an equation that we could employ the square root property on. That need not always be the case however. This is more than possible that we would require the quadratic formula to do some of these. We need to do an example of one of these just to make the point.