Can someone help me with these 3 algebra problems?

Question

1. g^2+2/3g+1/9=0
2.p^2-6/5p+9/25=0
3.(h+9)^2=3

Answer 1

1. The polynomial is a perfect square:

(g + 1/3)² = 0

g = -1/3

2. This polynomial is also a perfect square:

(p - 3/5)² = 0

p = 3/5

3. To do this one, you have to FOIL it out:

h² + 18h + 81 = 3

Subtract 3 from both sides:

h² + 18h + 78 = 0

You'll have to use the quadratic for this:

h = (-18 ± √(324 - 4(78)))/2
h = (-18 ± √12)/2
h = (-18 ± 2√3)/2
h = -9 ± √3

Puggy's way works also.

Answer 2

Answer 1:
g^2 +2/3 g +1/9 = 0
(g)^2 + 2 * 1/3 * g + (1/3)^2 = 0
(g + 1/3)^2 = 0
(g + 1/3)(g + 1/3) = 0
Obviously, g + 1/3 = 0
g = -1/3

Answer 2:
p^2 - 6/5 p + 9/25 = 0
(p)^2 - 2 * 3/5 * p + (3/5)^2 = 0
(p - 3/5)^2 = 0
(p - 3/5)(p - 3/5) = 0
Obviously,
p - 3/5 = 0
p = 3/5

Answer 3:
(h+9)^2 = 3
h + 9 = sqrt 3 ('sqrt' stands for 'square root of')
h = (sqrt 3) - 9

Answer 3

1. g^2 + (2/3)g + 1/9 = 0

This is actually a square binomial, and it factors into
[g + (1/3)]^2 = 0, which means if we take the square root of both sides,

g + (1/3) = 0, and
g = -1/3

2. p^2 - (6/5)p + 9/25 = 0

This is another binomial squared.

[p - (3/5)]^2 = 0, so
p - (3/5) = 0, and
p = 3/5

3. (h + 9)^2 = 3

Whenever you take the square root of both sides of an equation, you have to add a "plus or minus" symbol on the right hand side. Therefore, we take the square root of both sides to obtain

h + 9 = +/- sqrt(3)
h = -9 +/- sqrt(3)
h = {-9 + sqrt(3), -9 - sqrt(3)}

Answer 4

1. g^2+2/3g=-1/9
g*(g+2/3)=-1/9
g=-1/9/(g+2/3)

2. p^2-6/5p=-9/25
p*(p-6/5)=-9/25
p==9/25/(p-6/5)

3. h+9=3^-2
h=3^-2-9

Answer 5

3) You just take square root,
h+9 = ±√3
h = -9±√3

Answer 6

Uh-oh, not another one...
Are your variables on top of the fractions or on the bottom? i
If they're on top, you'd solve it the way Jim et al say, treating them as perfect square trinomials.
If they're on the bottom you'd have to treat this as a rational equation, multiply thru by the variable first to get it on top, then solve, with the stipulation that it cannot be 0.

Answer 7

Sorry I failed algebra.