what is the answer for y^2>=8y+9?

One way to solve this is to complete the square and to use absolute inequalities. As a reminder:

1) if |z| <= c (or |z| < c), it translates to an AND. It translates to

z <= c AND z >= -c (or z < c AND z > -c), also expressed as the compound inequality
-c <= z <= c (or -c < z < c)

2) if |z| >= c (or |z| > c), it translates to an OR. It translates to
(z >= c) OR (z <= -c) [ or (z > c) OR (z < -c).

y^2 >= 8y + 9
y^2 - 8y - 9 >= 0

Complete the square. Add 16 to both sides.

y^2 - 8y + 16 - 9 >= 16
(y - 4)^2 - 9 >= 16
(y - 4)^2 >= 25

sqrt( (y - 4)^2) = |y - 4|

|y - 4| >= 5

An absolute inequality that is greater than (or greater than or equal to) translates to an OR. It translates to

y - 4 >= 5 OR y - 4 <= -5

Solve each of those.

(y >= 9) OR (y <= -1)

sorry to the answers above mine but clearly the first one neglected the inequality sign, the second had problems with solving quadratic inequations

it's meant to be y^2>=8y+9
(y-9)(y+1) >=0
y >= 9 OR y<= -1
Which gives two separate sets of solutions, if you draw a parabola you can clearly see when y= f(x) = y^2 - 8y - 9 is larger than 0 (above x axis)

y^2>=8y+9

y^2-8y-9>=0
(y-9)(y+1)>=0

y-9>=0
y>=9

y+1>=0
y>=-1

y^2 - 8y - 9 = 0
( y - 9 ) (y + 1 ) = 0
y=9 or y = - 1

y^2 ≥ 8y + 9
y^2 - 8y - 9 ≥ 0
y^2 + y - 9y - 9 ≥ 0
(y^2 + y) - (9y + 9) ≥ 0
y(y + 1) - 9(y + 1) ≥ 0
(y + 1)(y - 9) ≥ 0

y + 1 ≥ 0
y ≤ -1

y - 9 ≥ 0
y ≥ 9

∴ y ≤ -1 or y ≥ 9

y² - 8y - 9 ≥ 0
(y - 9)(y + 1) ≥ 0

y - 9 ≥ 0 and y + 1 ≥ 0
y ≥ 9 and y ≥ - 1

y - 9 ≤ 0 and y + 1 ≤ 0
y ≤ 9 , y ≤ - 1

{ y : y ≤ - 1 U y ≥ 9 , y Є R }

right idea, moon light...

but when you get to (y - 9)(y + 1) >= 0
this is positive either when both factors are negative (y < -1) or when both are positive (y > 9)

so the answer is (-inf , -1] U [9 , inf)
or y <= -1 or y >= 9 (including -1 and 9, since the inequalities are greater than or =...)