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

2 Answers
Apr 24, 2017

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

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