Solution 3.2:3

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m (Lösning 3.2:3 moved to Solution 3.2:3: Robot: moved page)
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First, we move the
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<center> [[Image:3_2_3-1(2).gif]] </center>
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<math>\text{2}</math>
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to the right-hand side to get
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<math>\sqrt{3x-8}=x-2</math>
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<center> [[Image:3_2_3-2(2).gif]] </center>
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and then square away the root sign,
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(*)
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<math>3x-8=\left( x-2 \right)^{2}</math>
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or, with the right-hand side expanded
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<math>3x-8=x^{2}-4x+4</math>
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If we move over all the terms to the left-hand side, we get
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<math>x^{2}-7x+12=0</math>
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If we complete the square of the left-hand side,
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<math>\begin{align}
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& x^{2}-7x+12=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+12 \\
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& =\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{48}{4} \\
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& =\left( x-\frac{7}{2} \right)^{2}-\frac{1}{4} \\
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\end{align}</math>
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the equation can be written as
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<math>\left( x-\frac{7}{2} \right)^{2}=\frac{1}{4}</math>
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and the solutions are
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<math>x=\frac{7}{2}+\sqrt{\frac{1}{4}}=\frac{7}{2}+\frac{1}{2}=\frac{8}{2}=4</math>
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<math>x=\frac{7}{2}-\sqrt{\frac{1}{4}}=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3</math>
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To be on the safe side, we verify that
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<math>x=\text{3 }</math>
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and
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<math>x=\text{4}</math>
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satisfy the squared equation (*)
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<math>x=\text{3 }</math>: LHS
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<math>=3\centerdot 3-8=9-8=1</math>
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and RHS
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<math>=\left( 3-2 \right)^{2}=1</math>
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<math>x=\text{4}</math>: LHS
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<math>=3\centerdot 4-8=12-8=4</math>
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and RHS
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<math>=\left( 4\centerdot 2 \right)^{2}=4</math>
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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:
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<math>x=\text{3 }</math>: LHS
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<math>=\sqrt{3\centerdot 3-8}+2=\sqrt{9-8}+2=1+2=3</math>
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RHS
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<math>=3</math>
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<math>x=\text{4}</math>: LHS
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<math>=\sqrt{3\centerdot 4-8}-2=\sqrt{12-8}-2=2+2=4</math>
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RHS
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<math>=4</math>
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The solutions to the root equation are
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<math>x=\text{3 }</math>
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and
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<math>x=\text{4}</math>.

Revision as of 13:15, 23 September 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}.

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