t^2 + 4t + 11=0
2007-12-09 08:18:26 · 5 answers · asked by *TiNK* 3 in Science & Mathematics ➔ Mathematics
t² + 4t + 11=0 doesn't factor so use the quadratic solution:
t = ( -4 ± sqrt ( 4² - 4 ( 1 ) ( 11 ) ) / 2 )
t = ( -4 ± sqrt ( -28 ) ) / 2
Looks like both solutions are imaginary.
2007-12-09 08:24:02 · answer #1 · answered by jgoulden 7 · 1⤊ 0⤋
You have to use the quadratic equation becuase it doesnt factor. The quadratic equation is (-b± √ b²- 4ac) all over 2a.......
So, a= 1 ( becuase of the 1t²)
b= 4 (becuase of the 4t)
c= 11 (because of the ll in the equation)
So, you substitute those values into the equation.......
-4 ± √ 4² - 4 (1)(11) all over 2
Then, simplify it further
-4 ± √ 16 - 44 all over 2
-4± √ -28 all over 2
You can not get the answers ( atleast without inmaginary numbers)
2007-12-09 08:34:13 · answer #2 · answered by Lee K 2 · 0⤊ 0⤋
use the quadratic equation so t= -4(plus or minus) radical(16-(4)(1)(11) all over 2
Because the radical is negative, you cannot solve for t
2007-12-09 08:22:06 · answer #3 · answered by Ari R 3 · 0⤊ 0⤋
{-4 +/- ((16-44))^1/2}/2
={ -4 +/- (-28)^1/2}/2
= -1 +/-[ {(-28)^1/2}/2 ]
2007-12-09 08:25:10 · answer #4 · answered by vignesh.eems 2 · 0⤊ 0⤋
t = [- 4 ± √16 - 44) ] / 2
t = [- 4 ± √ (- 28) ] / 2
t = [- 4 ± i 2√(7) ] / 2
t = - 2 ± i √ 7
2007-12-09 10:14:52 · answer #5 · answered by Como 7 · 2⤊ 2⤋