How do you solve #0.4t^2+0.7t=0.3t-0.2# by completing the square?

1 Answer
Nov 27, 2017

Answer:

#t = -0.5 +- 0.5i #

Explanation:

#0.4t^2 + 0.7t = 0.3t -0.2#

Subtract #0.3t# from both sides

#0.4t^2 +0.4t = -0.2#

Divide the equation by the leading coefficient of the #t^2# term #0.4#.

#t^2 + t = -0.5#

To complete the square, divide the coefficient of the #t# term by #2#.

#t^2 + color(red)1t = -0.5#

#color(red)1 / 2 = color(blue)(0.5#

Square #color(blue)(0.5)# and add it to both sides of the equation.

#t^2 +t +(color(blue)(0.5))^2=-0.5 +(color(blue)(0.5))^2#

Simplify

#t^2 +t + 0.25 = -0.25#

Factor into a binomial squared. Note that the second term of the binomial is #color (blue)(0.25)#.

#(t+color(blue)(0.5))^2 = -0.25#

Square root both sides. Note that the negative sign in front of #-0.25# results in an #i#

#sqrt((t+color(blue)(0.5))^2) = sqrt(-0.25)#

#t + 0.5 = +-0.5i#

Subtract #0.5# from both sides. Place it in front of the #+-#.

#t=-0.5+-0.5i#