How do you solve the quadratic with complex numbers given #3t^2+8t+5=-2t^2#?

2 Answers
Mar 17, 2018

Answer:

#t=-4/5+3/5 i color(white)(..) and color(white)(..)t=-4/5-3/5 i#

Explanation:

Add #2t^2# to both sides

#5t^2+8t+5=0#

You can either complete the square or use the 'formula'

Keeping the standard form I will use #x#

#y=ax^2+bx+c =>x=(-b+-sqrt(b^2-4ac))/(2a)#

When I first came across these I made a point of writing the above at the start of every solution. It has stuck in my memory ever since. That was a long time ago so the method works!

#a=+5;b=+8 and c=+5# giving, in this case #t#:

#t=(-8+-sqrt(8^2-4(5)(5)))/(2(5))#

#t=-8/10+-sqrt(-36)/10#

Bit #-36# is the same as #36xx(-1)#

#t=-8/10+- [(sqrt(36)color(white)(.) sqrt(-1))/10] #

But #sqrt(36)=6 and sqrt(-1) = i#

#t=-4/5+-3/5 color(white)(.) i#

#t=-4/5+3/5 i color(white)(..) and color(white)(..)t=-4/5-3/5 i#

Mar 17, 2018

Answer:

#t_1=-4/5+i3/5 or t_2=-4/5-i3/5#

Explanation:

#3t^2+8t+5=-2t^2|+2t^2#
#5t^2+8t+5=0|:5#
#t^2+8/5t+1=0#

Completing the square...

#(t+4/5)^2+1-16/25=0|-9/25|sqrt()#
#t+4/5=+-sqrt(-9/25)|-4/5#
#t=-4/5+-sqrt(-9/25)#

Determinating the complex root...

#sqrt(-a)=i*sqrt(a)#

#+-sqrt(-9/25)=+-i3/5#
#t_1=-4/5+i3/5 or t_2=-4/5-i3/5#