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1.1 Lösning 4a - Versionshistorik
2026-06-29T22:11:54Z
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Taifun: Created page with "<math>\begin{align} \sqrt{x^2 + 1} & = x - 3 & & \qquad | \; (\;\;\;)^2 \\ x^2 + 1 & = (x - 3)^2 & & \\ ..."
2010-11-21T09:17:00Z
<p>Created page with "<math>\begin{align} \sqrt{x^2 + 1} & = x - 3 & & \qquad | \; (\;\;\;)^2 \\ x^2 + 1 & = (x - 3)^2 & & \\ ..."</p>
<p><b>Ny sida</b></p><div><math>\begin{align} \sqrt{x^2 + 1} & = x - 3 & & \qquad | \; (\;\;\;)^2 \\<br />
x^2 + 1 & = (x - 3)^2 & & \\<br />
x^2 + 1 & = x^2 - 6\,x + 9 & & \qquad | \;\; - x^2 \\<br />
1 & = - 6\,x + 9 & & \qquad | \;\; + 6\,x \\<br />
6\,x + 1 & = 9 & & \qquad | \;\; - 1 \\<br />
6\, x & = 8 & & \qquad | \;\; / 6 \\<br />
x & = {8 \over 6} = {4 \over 3} \\<br />
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\end{align}</math><br />
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Prövning:<br />
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VL: <math> \sqrt{\left({4 \over 3}\right)^2 + 1} = \sqrt{{16 \over 9} + 1} = \sqrt{{16 \over 9} + {9 \over 9}} = \sqrt{{25 \over 9}} = {5 \over 3} </math><br />
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HL: <math> {4 \over 3} - 3 = {4 \over 3} - {9 \over 3} = - {5 \over 3} </math><br />
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VL <math> \not= </math> HL <math> \Rightarrow\, x = {4 \over 3} </math> är en falsk rot.</div>
Taifun