How do you solve 1/t^2 + 8/t + 15 = 0 The answers are -1/5 and -1/3?

Answer 1

1/t^2 + 8/t + 15 = 0 Let's mutiply each term by t^2 to make it look more normal

15t^2 +8t +1=0

so by inspection

(5t+1)(3t+1)=0

thus t=-1/15 and -1/3

HTH :-)

Answer 2

1/t^2 + 8/t + 15 = 0

multiply t^2 for both sides
1 + 8t + 15t^2 = 0

subtract 1 for both sides
15t^2 + 8t = -1

divide 15 for both sides
t^2 + (8/15)t = -1/15

add (b/2)^2 for both sides
t^2 + (8/15)t + 16/225 = 1/225

(t + 4/15)^2 = 1/225

take a square root
t + 4/15 = +/- 1/15

subtract 4/15 for both sides
t = -4/15 +/- 1/15

t = -4/15 - 1/15 = -1/5
t = -4/15 + 1/15 = -1/3

Answer 3

1/t^2 + 8/t + 15 = 0
=> 1 + 8t + 15t^2 = 0 ......X by LCM t^2

=> ( 1+ 5t )( 1 + 3t ) = 0
=> 1+ 5t = 0 OR 1 + 3t = 0
=> t = - 1/5 OR t = -1/3

QED

Answer 4

1/t^2 + 8/t + 15 = 0 = (1/t + 3) (1/t+5)

roots are

t = -1/3 and -1/5

Answer 5

Don't let "t" in the denominator fake you out. In fact, if you substitute T=1/t in, you get
T^2+8T+15=0.
Just factor this (pretty easy), solve for T, and then for t.

Answer 6

u use the quadratic formula
x = (-b(+or-)(b^2 - 4ac)^1/2)/2a
(look it up for a better understanding)

and then
a=1
b=8
c=15