From Förberedande kurs i matematik 1

Revision as of 09:18, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.2:3 moved to Solution 3.2:3: Robot: moved page) ← Previous diff		Current revision (12:06, 1 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	First, we move the 2 to the right-hand side to get <math>\sqrt{3x-8}=x-2</math> and then square away the root sign,
-	<center> [[Image:3_2_3-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>3x-8 = (x-2)^{2}</math>|(*)}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:3_2_3-2(2).gif]] </center>	+	or, with the right-hand side expanded
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>3x-8=x^{2}-4x+4\,\textrm{.}</math>}}
		+
		+	If we move over all the terms to the left-hand side, we get
		+
		+	{{Displayed math||<math>x^{2}-7x+12=0\,\textrm{.}</math>}}
		+
		+	If we complete the square of the left-hand side,
		+
		+	{{Displayed math||<math>\begin{align}
		+	x^2-7x+12 &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2 + 12\\[5pt]
		+	&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{49}{4} + \frac{48}{4}\\[5pt]
		+	&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{1}{4}
		+	\end{align}</math>}}
		+
		+	the equation can be written as
		+
		+	{{Displayed math||<math>\Bigl(x-\frac{7}{2}\Bigr)^{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\,\textrm{.}</math>
		+
		+	To be on the safe side, we verify that <math>x=3</math> and <math>x=4</math> satisfy the squared equation (*)
		+
		+	{|
		+	||<ul><li>''x''&nbsp;=&nbsp;3:</li></ul>
		+	||<math>\ \text{LHS} = 3\cdot 3-8 = 9-8 = 1</math> and
		+	|-
		+	||
		+	||<math>\ \text{RHS} = (3-2)^2 = 1</math>
		+	|-
		+	||<ul><li>''x''&nbsp;=&nbsp;4:</li></ul>
		+	||<math>\ \text{LHS} = 3\cdot 4-8 = 12-8 = 4</math> and
		+	|-
		+	||
		+	||<math>\ \text{RHS} = (4-2)^2 = 4</math>
		+	|}
		+
		+	Because we squared the root equation, possible spurious roots turn up and we therefore have to verify the solutions when we go back to the original root equation:
		+
		+	{|
		+	||<ul><li>''x''&nbsp;=&nbsp;3:</li></ul>
		+	||<math>\ \text{LHS} = \sqrt{3\cdot 3-8} + 2 = \sqrt{9-8} + 2 = 1+2 = 3</math> and
		+	|-
		+	||
		+	||<math>\ \text{RHS} = 3</math>
		+	|-
		+	||<ul><li>''x''&nbsp;=&nbsp;4:</li></ul>
		+	||<math>\ \text{LHS} = \sqrt{3\cdot 4-8}-2 = \sqrt{12-8}+2 = 2+2 = 4</math> and
		+	|-
		+	||
		+	||<math>\ \text{RHS} = 4</math>
		+	|}
		+
		+	The solutions to the root equation are <math>x=3</math> and <math>x=4</math>.

---

Current revision

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

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

(*)

or, with the right-hand side expanded

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

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

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

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

\displaystyle \begin{align} x^2-7x+12 &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2 + 12\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{49}{4} + \frac{48}{4}\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{1}{4} \end{align}

the equation can be written as

\displaystyle \Bigl(x-\frac{7}{2}\Bigr)^{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\,\textrm{.}

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

x = 3:	\displaystyle \ \text{LHS} = 3\cdot 3-8 = 9-8 = 1 and
	\displaystyle \ \text{RHS} = (3-2)^2 = 1
x = 4:	\displaystyle \ \text{LHS} = 3\cdot 4-8 = 12-8 = 4 and
	\displaystyle \ \text{RHS} = (4-2)^2 = 4

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

x = 3:	\displaystyle \ \text{LHS} = \sqrt{3\cdot 3-8} + 2 = \sqrt{9-8} + 2 = 1+2 = 3 and
	\displaystyle \ \text{RHS} = 3
x = 4:	\displaystyle \ \text{LHS} = \sqrt{3\cdot 4-8}-2 = \sqrt{12-8}+2 = 2+2 = 4 and
	\displaystyle \ \text{RHS} = 4

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