Answer: number 1 2 and four are correct
Step-by-step explanation:
I DID THE SAME THING
I'm guessing your problem is this:
y³ - 9y² + y - 9 = 0
right?
In solving this problem, I recommend doing this:
y³ - 9y² + y - 9 = 0
Factor out a y² from the first two numbers in the problem:
y²(y - 9) + (y - 9) = 0
Separate the parentheses which means y - 9 goes on one side. The y² added a one since it came from the + 1 in the middle of expression. When you're separating parentheses like this you just take the outside numbers and combine them together. Since + 1 came from the outside of the (y - 9) and y² also was sitting on the outside of (y - 9) combine them to make y² + 1. Like this:
(y² + 1)(y - 9) = 0
Now separate your two parentheses to two separate problems:
(y² + 1) = 0 and (y - 9) = 0
Now you're y² + 1 will equal:
y² = -1
y = √-1 <-- This number doesn't exist so it will be an imaginary number (i). If you guys didn't learn that in your class I recommend just leaving it as i for that part.
Now solve y - 9 = 0:
y = 9 <-- Since we added nine to both sides to get this.
So you're final answer should be y = i and 9
Rearrange to: x^2 + 18x - 25 = 0
Then factorise (I like to do it by completing the square):
(x + 9)^2 -81 -25 = 0
(x + 9)^2 = 106
x + 9 = sqrt(106)
The sol'ns are:
x = -9 + and - sqrt(106)
