How do you solve #sqrt(1-2t)=1+t# and check your solution?

2 Answers
Apr 24, 2017

Answer:

#t=0#

Explanation:

#sqrt(1-2t)=1+t#

Square both sides to get rid of the square root.

#(sqrt(1-2t))^2=(1+t)^2#

#1-2t=t^2+2t+1#

Recognize that we should move everything to the right side of the equals sign to get a complete quadratic function for this equation.

#1-2t=t^2+2t+1#

#0=t^2+2tcancel(+1)cancel(-1)+2t#

#0=t^2+4t#

Take out common factor #t#

#0=t(t+4)#

Now we have our two values for #t# :

#t=0#
#t=-4# #<--- #Remember it is not positive 4 because we have not put the 4 on the other side of our equals sign yet

Now to check our possible solutions we substitute our values for #t# into the ORIGINAL EQUATION

Lets start with #-4#

#sqrt(1-2(-4))=1+(-4)#

#sqrt(1+8)=1-4#

#sqrt(9)=-3#

#3=-3#
#color(red)"So -4 is NOT an answer for this equation"##", lets try 0"#

#sqrt(1-2(0))=1+(0)#

#sqrt(1-0)=1+0#

#sqrt(1)=1#

#1=1#

#color(green)"0 IS an answer for this equation"#

Apr 24, 2017

Answer:

#color(blue)(t=0#

Explanation:

#sqrt(1-2t)=1+t#

Square L.H.S. and R.H.S

#:.(sqrt(1-2t)^2)=(1+t)^2#

#:.sqrt(1-2t)*sqrt(1-2t)=1-2t#

#:.(1+t)*(1+t)=1+2t+t^2#

#:.1+2t+t^2=1-2t#

#:.t^2=1-1-2t-2t#

#:.t^2=-4t#

#:.t^2+4t=0#

#t(t+4)=0#

#color(blue)(t=0# or # color(blue)(t=-4#

substitute #color(blue)(t=-4#

#:.sqrt(1-2(color(blue)(-4)))=1+(color(blue)(-4))#

#:.sqrt(1+8)=1-4#

#:.+-sqrt(9)=-3#

#-3# or #+3 =-3#

#:.-3=-3#

#:.3=-3# extraneous solution

substitute #color(blue)(t=0#

#:.sqrt(1-2(color(blue)(0)))=1+(color(blue)(0))#

#:.sqrt(1)=1#

#:.1=1#