SOLUTION: Give exact and approximate solutions to three decimal places
y^2-20y+100=16
What are the exact solutions?
y=?
(Type an exact answer as needed)
(Rationalize all denominators)
Question 626391: Give exact and approximate solutions to three decimal places
y^2-20y+100=16
What are the exact solutions?
y=?
(Type an exact answer as needed)
(Rationalize all denominators)
(express complex numbers in terms of I)
(use a comma to separate answers as needed)
There are no approximate solutions since the solutions are integers
Does this mean that:
1)y^2 is the ^2 of y
2)+100=10^2
3)ignoring the sign (-) on -20, the product of square roots of 1st and 3rd term = 2*y*10 (20y)
4)y = (y-10)^2=16
y-10= ±4
y = 10 ±5
Am I doing this right? Answer by solver91311(24713) (Show Source):
Don't try to do anything until your equation is in standard form, which is to say all of it in the LHS and the RHS is zero. In other words, it needs to look like:
So, in your case, add -16 to both sides:
Now you can factor and use the Zero Product Rule. Note -6 times -14 is 84 and -6 plus -14 is -20.
John
My calculator said it, I believe it, that settles it
