1.1 Lösning 4a

$\begin{align} \sqrt{x^2 + 1} & = x - 3 & & \qquad | \; (\;\;\;)^2 \\ x^2 + 1 & = (x - 3)^2 & & \\ x^2 + 1 & = x^2 - 6\,x + 9 & & \qquad | \;\; - x^2 \\ 1 & = - 6\,x + 9 & & \qquad | \;\; + 6\,x \\ 6\,x + 1 & = 9 & & \qquad | \;\; - 1 \\ 6\, x & = 8 & & \qquad | \;\; / 6 \\ x & = {8 \over 6} = {4 \over 3} \\ \end{align}$

Prövning:

VL $$\sqrt{\left({4 \over 3}\right)^2 + 1} = \sqrt{{16 \over 9} + 1} = \sqrt{{16 \over 9} + {9 \over 9}} = \sqrt[[:Mall:25 \over 9]] = {5 \over 3}$$

HL $${4 \over 3} - 3 = {4 \over 3} - {9 \over 3} = - {5 \over 3}$$

VL $\not=$ HL $\Rightarrow\, x = {4 \over 3}$ är en falsk rot.