Solution 3.2:3
From Förberedande kurs i matematik 1
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| - | First, we move the | + | 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, |
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| - | to the right-hand side to get | + | |
| - | <math>\sqrt{3x-8}=x-2</math> | + | |
| - | and then square away the root sign, | + | |
| - | + | {{Displayed math||<math>3x-8 = (x-2)^{2}</math>|(*)}} | |
| - | <math>3x-8= | + | |
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or, with the right-hand side expanded | or, with the right-hand side expanded | ||
| - | + | {{Displayed math||<math>3x-8=x^{2}-4x+4\,\textrm{.}</math>}} | |
| - | <math>3x-8=x^{2}-4x+4</math> | + | |
| - | + | ||
If we move over all the terms to the left-hand side, we get | If we move over all the terms to the left-hand side, we get | ||
| - | + | {{Displayed math||<math>x^{2}-7x+12=0\,\textrm{.}</math>}} | |
| - | <math>x^{2}-7x+12=0</math> | + | |
| - | + | ||
If we complete the square of the left-hand side, | If we complete the square of the left-hand side, | ||
| - | + | {{Displayed math||<math>\begin{align} | |
| - | <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>}} |
| - | \end{align}</math> | + | |
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the equation can be written as | the equation can be written as | ||
| - | + | {{Displayed math||<math>\Bigl(x-\frac{7}{2}\Bigr)^{2} = \frac{1}{4}</math>}} | |
| - | <math>\ | + | |
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and the solutions are | 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> | |
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| - | <math>x=\frac{7}{2}-\sqrt{\frac{1}{4}}=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3 | + | |
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| - | <math>x= | + | To be on the safe side, we verify that <math>x=3</math> and <math>x=4</math> satisfy the squared equation (*) |
| - | <math>= | + | |
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| + | {| | ||
| + | ||<ul><li>''x'' = 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'' = 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: | |
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| - | + | ||
| - | + | ||
| + | {| | ||
| + | ||<ul><li>''x'' = 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'' = 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 | + | The solutions to the root equation are <math>x=3</math> and <math>x=4</math>. |
| - | <math>x= | + | |
| - | and | + | |
| - | <math>x= | + | |
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 (*)
| \displaystyle \ \text{LHS} = 3\cdot 3-8 = 9-8 = 1 and |
| \displaystyle \ \text{RHS} = (3-2)^2 = 1 | |
| \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:
| \displaystyle \ \text{LHS} = \sqrt{3\cdot 3-8} + 2 = \sqrt{9-8} + 2 = 1+2 = 3 and |
| \displaystyle \ \text{RHS} = 3 | |
| \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.
