1.1 Lösning 13
\(\begin{align} 6\;x & = 1 - \sqrt{36\;x^2 - {1 \over x}} & & \qquad | - 1 \\ 6\;x - 1 & = - \sqrt{36\;x^2 - {1 \over x}} & & \qquad | \; (\;\;\;)^2 \\ (6\,x - 1)^2 & = 36\,x^2 - {1 \over x} \\ 36\,x^2 - 12\,x + 1 & = 36\,x^2 - {1 \over x} & & \qquad | - 36\,x^2 \\ - 12\,x + 1 & = - {1 \over x} & & \qquad | \cdot x \\ - 12\,x^2 + x & = - 1 & & \qquad \\ 12\,x^2 - x - 1 & = 0 & & \qquad | \; /\;12 \\ x^2 - {1 \over 12}\,x - {1 \over 12} & = 0 \\ x_{1,2} & = {1 \over 24} \pm \sqrt{{1 \over 24^2} + {1 \over 12}} \\ x_{1,2} & = {1 \over 24} \pm \sqrt{{1 \over 24^2} + {4\cdot 12 \over 24^2}} \\ x_{1,2} & = {1 \over 24} \pm \sqrt[[:Mall:49 \over 24^2]] \\ x_{1,2} & = {1 \over 24} \pm {7 \over 24} \\ x_{1,2} & = 15 \pm 6 \\ x_1 & = 21 \\ x_2 & = 9 \\ \end{align}\)
Prövning av \( x_1 = 21\, \):
VL\[ \sqrt{21 + 2 + \sqrt{2 \cdot 21 + 7}} = \sqrt{23 + \sqrt{2 \cdot 21 + 7}} = \sqrt{23 + \sqrt{42 + 7}} = \]
\( = \sqrt{23 + \sqrt{49}} = \sqrt{23 + 7} = \sqrt{30} = 5,48 \)
HL\[ 4\, \]
VL \(\not=\) HL \( \Rightarrow\, x_1 = 21 \) är en falsk rot.
Prövning av \( x_2 = 9\, \):
VL\[ \sqrt{9 + 2 + \sqrt{2 \cdot 9 + 7}} = \sqrt{11 + \sqrt{2 \cdot 9 + 7}} = \sqrt{11 + \sqrt{18 + 7}} = \]
\( = \sqrt{11 + \sqrt{25}} = \sqrt{11 + 5} = \sqrt{16} = 4 \)
HL\[ 4\, \]
VL = HL \( \Rightarrow\, x_2 = 9 \) är en sann rot.
Svar\[ x = 9\, \] är rotekvationens enda lösning.
