Solve. 3t^ 2 + 18^t + 3 = 0?

Question

Solve. 3t^ 2 + 18^t + 3 = 0

t = (smaller value)

t = (larger value)

Answer 1

t ² + 6 t + 1 = 0
t = [- 6 ± √(32)] / 2
t = [- 6 ± 4√(2)] / 2
t = - 3 ± 2√2

Answer 2

Is your equation correct? or did you mean 3t^2+18t+3=0 because right now you have 18^t, which I don't know how to solve. If you meant 3t^2+18t+3=0 , then I would solve it using the completing the square method

1. The coefficient of the squared term needs to be 1...to do this, divide both sides by 3

3t^2+18t+3=0
------------------
3

t^2+6t+1=0

2. Now isolate the variables from the constant...to do this, subtract 1 from both sides

t^2+6t+1-1=0-1
t^2+6t=-1

3. Now you must complete the square by creating a perfect squared trinomial...to do this, take the coefficient of the non-squared variable divide by 2, then raise to the second power. (sounds complicated, but it's not. The non-squared term in this case is 6t, therefore the coefficient of this term is 6.) Be sure to add this new term to both sides...

t^2+6t+ (6/2)^2=-1+ (6/2)^2
t^2+6t+9= -1+9 (do you see where I got 9 from...6/2=3, then 3^2=9)

4. Simplify by factoring...

t^2+6t+9=-1+9
(t+3)^2=8

5. Take the square root of both sides to get rid of the exponent

sqrt(t+3)^2= +/-sqrt 8
t+3=+/-sqrt 8

6. Now solve for t...to do this subtract 3 from both sides

t=-3+/-sqrt 8

note: this is two answers
t= -3 + sqrt 8 or t= -3 - sqrt 8

Hope this helps...

Answer 3

3t^ 2 + 18^t + 3 = 0
t^2+6t+1=0
t^2+6t+9=1+9
(t+3)^2=10
t=+/-sqrt(10)-3

Answer 4

divide through by 3:

1/3 ( 3t^2 +18t +3 =0)
t^2 +6t +3 =0

now do the quadratic formula:
[-b +/- sqrt (b^2 -4ac)]/2a

[-6 +/- sqrt (36 -(4)(3) )] /2 =x

t= -3 +/- [sqrt (36-12) ]/2
t= -3 +/- [sqrt (24) ]/2
t= -3 +/- 2 [sqrt (6) ]/2
t= -3 +/- sqrt (6)

t= -3 + sqrt (6)
t= -3 - sqrt (6)

these are the two solutions to the equation.

Answer 5

Factor 3(t^2-6t+5)>0 3(t-1)(t-5)>0 so for a product of two numbes to be bigger then 0, either both numbers are positive or both are negative. so if both numbers are positive, t-1>0 and t-5>0 which means t>1 and t>5, but since 5>1, then saying t>5 covers the t>1 case. now if both are negative that also will give a product >0, so t-1<0 and t-5<0 which means t<1 and t<5, but if t<1 that covers the t<5 case, so you use the t<1. so either t<1 or t>5. solution is (-infinite,1) U (5,infinite)

Answer 6

3t^2+18t+3=0
3(t^2+6t+1)=0
t^2+6t+1=0 .....division by 3
t={-b+/-rt(b^2-4ac} / 2a
a=1,b=6,c=1
t={-6+/-rt(36-4(1)(1))}/2
t={-6+/-rt32}/2
t={-6+/-4rt2}/2
t=2{-3+/-2rt2}/2
t=-3+2rt2 or -3-3rt2
rt2 is 1.4142
t=-3+2(1.4142), or -3-2(1.4142)
t= -0.1716 or -5.8284