Aus Online Mathematik Brückenkurs 1

Version vom 09:18, 9. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 3.2:3 moved to Solution 3.2:3: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 13:15, 23. Sep. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	First, we move the
-	<center> [[Image:3_2_3-1(2).gif]] </center>	+	<math>\text{2}</math>
-	{{NAVCONTENT_STOP}}	+	to the right-hand side to get
-	{{NAVCONTENT_START}}	+	<math>\sqrt{3x-8}=x-2</math>
-	<center> [[Image:3_2_3-2(2).gif]] </center>	+	and then square away the root sign,
-	{{NAVCONTENT_STOP}}	+
		+	(*)
		+	<math>3x-8=\left( x-2 \right)^{2}</math>
		+
		+
		+	or, with the right-hand side expanded
		+
		+
		+	<math>3x-8=x^{2}-4x+4</math>
		+
		+
		+	If we move over all the terms to the left-hand side, we get
		+
		+
		+	<math>x^{2}-7x+12=0</math>
		+
		+
		+	If we complete the square of the left-hand side,
		+
		+
		+	<math>\begin{align}
		+	& x^{2}-7x+12=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+12 \\
		+	& =\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{48}{4} \\
		+	& =\left( x-\frac{7}{2} \right)^{2}-\frac{1}{4} \\
		+	\end{align}</math>
		+
		+
		+	the equation can be written as
		+
		+
		+	<math>\left( x-\frac{7}{2} \right)^{2}=\frac{1}{4}</math>
		+
		+
		+	and the solutions are
		+
		+
		+	<math>x=\frac{7}{2}+\sqrt{\frac{1}{4}}=\frac{7}{2}+\frac{1}{2}=\frac{8}{2}=4</math>
		+
		+
		+	<math>x=\frac{7}{2}-\sqrt{\frac{1}{4}}=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3</math>
		+
		+	To be on the safe side, we verify that
		+	<math>x=\text{3 }</math>
		+	and
		+	<math>x=\text{4}</math>
		+	satisfy the squared equation (*)
		+
		+
		+	<math>x=\text{3 }</math>: LHS
		+	<math>=3\centerdot 3-8=9-8=1</math>
		+	and RHS
		+	<math>=\left( 3-2 \right)^{2}=1</math>
		+
		+
		+	<math>x=\text{4}</math>: LHS
		+	<math>=3\centerdot 4-8=12-8=4</math>
		+	and RHS
		+	<math>=\left( 4\centerdot 2 \right)^{2}=4</math>
		+
		+
		+	Because we squared the root equation, possible false roots turn up and we therefore have to verify the solutions when we go back to the original root equation:
		+
		+
		+	<math>x=\text{3 }</math>: LHS
		+	<math>=\sqrt{3\centerdot 3-8}+2=\sqrt{9-8}+2=1+2=3</math>
		+
		+	RHS
		+	<math>=3</math>
		+
		+
		+	<math>x=\text{4}</math>: LHS
		+	<math>=\sqrt{3\centerdot 4-8}-2=\sqrt{12-8}-2=2+2=4</math>
		+
		+	RHS
		+	<math>=4</math>
		+
		+
		+	The solutions to the root equation are
		+	<math>x=\text{3 }</math>
		+	and
		+	<math>x=\text{4}</math>.

---

Version vom 13:15, 23. Sep. 2008

First, we move the \displaystyle \text{2} to the right-hand side to get \displaystyle \sqrt{3x-8}=x-2 and then square away the root sign,

(*) \displaystyle 3x-8=\left( x-2 \right)^{2}

or, with the right-hand side expanded

\displaystyle 3x-8=x^{2}-4x+4

If we move over all the terms to the left-hand side, we get

\displaystyle x^{2}-7x+12=0

If we complete the square of the left-hand side,

\displaystyle \begin{align} & x^{2}-7x+12=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+12 \\ & =\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{48}{4} \\ & =\left( x-\frac{7}{2} \right)^{2}-\frac{1}{4} \\ \end{align}

the equation can be written as

\displaystyle \left( x-\frac{7}{2} \right)^{2}=\frac{1}{4}

and the solutions are

\displaystyle x=\frac{7}{2}+\sqrt{\frac{1}{4}}=\frac{7}{2}+\frac{1}{2}=\frac{8}{2}=4

\displaystyle x=\frac{7}{2}-\sqrt{\frac{1}{4}}=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3

To be on the safe side, we verify that \displaystyle x=\text{3 } and \displaystyle x=\text{4} satisfy the squared equation (*)

\displaystyle x=\text{3 }: LHS \displaystyle =3\centerdot 3-8=9-8=1 and RHS \displaystyle =\left( 3-2 \right)^{2}=1

\displaystyle x=\text{4}: LHS \displaystyle =3\centerdot 4-8=12-8=4 and RHS \displaystyle =\left( 4\centerdot 2 \right)^{2}=4

Because we squared the root equation, possible false roots turn up and we therefore have to verify the solutions when we go back to the original root equation:

\displaystyle x=\text{3 }: LHS \displaystyle =\sqrt{3\centerdot 3-8}+2=\sqrt{9-8}+2=1+2=3

RHS \displaystyle =3

\displaystyle x=\text{4}: LHS \displaystyle =\sqrt{3\centerdot 4-8}-2=\sqrt{12-8}-2=2+2=4

RHS \displaystyle =4

The solutions to the root equation are \displaystyle x=\text{3 } and \displaystyle x=\text{4}.