Solve the following equation for t:

by **blu3fan** » Sun Jan 28, 2007 12:25 pm

by **keentoknow** » Sat May 05, 2007 1:08 pm

muliply by 25-t² but remember that t≠±5 to obtain
9(5+t)-4(5-t)+1=0 or
45+9t-20+4t+1-0 or
13t-26=0 so t=2

by **MV RAO** » Wed Dec 09, 2009 7:42 am

keentoknow wrote:muliply by 25-t² but remember that t≠±5 to obtain
9(5+t)-4(5-t)+1=0 or
45+9t-20+4t+1-0 or
13t-26=0 so t=2

t=2 is wrong. It is 13t + 26 = 0 or t = -2. Put back value of t and check!

by **Kazuya07** » Thu Feb 19, 2015 6:24 am

16-4x^2>0
-3x^2<0
x^2-2x-3<0

(x+1)^2>3(x+2)+1
(x-1)^2+x(1-x)>2+(1-x)(x+1)
(1-x)^2<=1

DdD DeViL

by **Kazuya07** » Thu Mar 05, 2015 6:08 am

16-4x^2>0
-3x^2<0
x^2-2x-3<0

(x+1)^2>3(x+2)+1
(x-1)^2+x(1-x)>2+(1-x)(x+1)
(1-x)^2<=1

_________________________

by **Guest** » Sat Mar 07, 2015 6:57 am

9(5+t)-4(5-t)+1=0
45+9t-20+4t+1=0
13t+26=0
13t = -26
t = -2