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

Apr 24, 2017

$t = 0$

#### Explanation:

$\sqrt{1 - 2 t} = 1 + t$

Square both sides to get rid of the square root.

${\left(\sqrt{1 - 2 t}\right)}^{2} = {\left(1 + t\right)}^{2}$

$1 - 2 t = {t}^{2} + 2 t + 1$

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

$1 - 2 t = {t}^{2} + 2 t + 1$

$0 = {t}^{2} + 2 t \cancel{+ 1} \cancel{- 1} + 2 t$

$0 = {t}^{2} + 4 t$

Take out common factor $t$

$0 = t \left(t + 4\right)$

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 $- 4$

$\sqrt{1 - 2 \left(- 4\right)} = 1 + \left(- 4\right)$

$\sqrt{1 + 8} = 1 - 4$

$\sqrt{9} = - 3$

$3 = - 3$
$\textcolor{red}{\text{So -4 is NOT an answer for this equation}}$$\text{, lets try 0}$

$\sqrt{1 - 2 \left(0\right)} = 1 + \left(0\right)$

$\sqrt{1 - 0} = 1 + 0$

$\sqrt{1} = 1$

$1 = 1$

$\textcolor{g r e e n}{\text{0 IS an answer for this equation}}$

Apr 24, 2017

color(blue)(t=0

#### Explanation:

$\sqrt{1 - 2 t} = 1 + t$

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

$\therefore \left({\sqrt{1 - 2 t}}^{2}\right) = {\left(1 + t\right)}^{2}$

$\therefore \sqrt{1 - 2 t} \cdot \sqrt{1 - 2 t} = 1 - 2 t$

$\therefore \left(1 + t\right) \cdot \left(1 + t\right) = 1 + 2 t + {t}^{2}$

$\therefore 1 + 2 t + {t}^{2} = 1 - 2 t$

$\therefore {t}^{2} = 1 - 1 - 2 t - 2 t$

$\therefore {t}^{2} = - 4 t$

$\therefore {t}^{2} + 4 t = 0$

$t \left(t + 4\right) = 0$

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

substitute color(blue)(t=-4

$\therefore \sqrt{1 - 2 \left(\textcolor{b l u e}{- 4}\right)} = 1 + \left(\textcolor{b l u e}{- 4}\right)$

$\therefore \sqrt{1 + 8} = 1 - 4$

$\therefore \pm \sqrt{9} = - 3$

$- 3$ or $+ 3 = - 3$

$\therefore - 3 = - 3$

$\therefore 3 = - 3$ extraneous solution

substitute color(blue)(t=0

$\therefore \sqrt{1 - 2 \left(\textcolor{b l u e}{0}\right)} = 1 + \left(\textcolor{b l u e}{0}\right)$

$\therefore \sqrt{1} = 1$

$\therefore 1 = 1$