1.1 Lösning 13

$\begin{align} 6\;x & = 1 - \sqrt{36\;x^2 - {1 \over x}} & & \qquad | - 1 \\ 6\;x - 1 & = - \sqrt{36\;x^2 - {1 \over x}} & & \qquad | \; (\;\;\;)^2 \\ (6\,x - 1)^2 & = 36\,x^2 - {1 \over x} \\ 36\,x^2 - 12\,x + 1 & = 36\,x^2 - {1 \over x} & & \qquad | - 36\,x^2 \\ - 12\,x + 1 & = - {1 \over x} & & \qquad | \cdot x \\ - 12\,x^2 + x & = - 1 & & \qquad \\ 12\,x^2 - x - 1 & = 0 & & \qquad | \; /\;12 \\ x^2 - {1 \over 12}\,x - {1 \over 12} & = 0 \\ x_{1,2} & = {1 \over 24} \pm \sqrt{{1 \over 24^2} + {1 \over 12}} \\ x_{1,2} & = {1 \over 24} \pm \sqrt{{1 \over 24^2} + {4\cdot 12 \over 24^2}} \\ x_{1,2} & = {1 \over 24} \pm \sqrt[[:Mall:49 \over 24^2]] \\ x_{1,2} & = {1 \over 24} \pm {7 \over 24} \\ x_1 & = {1 \over 3} \\ x_2 & = -{1 \over 4} \\ \end{align}$

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Prövning av $x_1 = {1 \over 3}$:

VL $$6\cdot {1 \over 3} = 2$$

HL $$1 - \sqrt{36\cdot \left({1 \over 3}\right)^2 - {1 \over {1 \over 3}}} = 1 - \sqrt{36\cdot {1 \over 9} - 3} = 1 - \sqrt{4 - {3}} = 1 - 1 = 0$$

VL $\not=$ HL $\Rightarrow\, x_1 = {1 \over 3}$ är en falsk rot.

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Prövning av $x_2 = -{1 \over 4}$:

VL $$6\cdot \left(-{1 \over 4}\right) = -{3 \over 2}$$

HL $$1 - \sqrt{36\cdot \left(-{1 \over 4}\right)^2 - {1 \over -{1 \over 4}}} = 1 - \sqrt{36\cdot {1 \over 16} + 4} = 1 - \sqrt{{9\over4} + 4} =$$

$= 1 - \sqrt{{9\over4} + {16\over4}} = 1 - \sqrt[[:Mall:25\over4]] = 1 - {5\over2} = {2\over2} - {5\over2} = -{3 \over 2}$

VL = HL $\Rightarrow\, x_2 = -{1 \over 4}$ är en sann rot.

Svar $$x = -{1 \over 4}$$ är rotekvationens enda lösning.