Solution 2.3:2f
From Förberedande kurs i matematik 1
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- | {{ | + | We divide both sides by 3 and complete the square on the left-hand side, |
- | + | ||
- | {{ | + | {{Displayed math||<math>\begin{align} |
+ | x^{2}-\frac{10}{3}x+\frac{8}{3} | ||
+ | &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \Bigl(\frac{5}{3}\Bigr)^{2} + \frac{8}{3}\\[5pt] | ||
+ | &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{25}{9} + \frac{24}{9}\\[5pt] | ||
+ | &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{1}{9}\,\textrm{.} | ||
+ | \end{align}</math>}} | ||
+ | |||
+ | The equation then becomes | ||
+ | |||
+ | {{Displayed math||<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,}</math>}} | ||
+ | |||
+ | and taking the square root gives the solutions as | ||
+ | |||
+ | :*<math>x-\tfrac{5}{3} = \sqrt{\tfrac{1}{9}} = \tfrac{1}{3}\,,\quad</math> i.e. <math>x = \tfrac{5}{3} + \tfrac{1}{3} = \tfrac{6}{3} = 2\,\textrm{,}</math> | ||
+ | |||
+ | :*<math>x-\tfrac{5}{3} = -\sqrt{\tfrac{1}{9}} = -\tfrac{1}{3}\,,\quad</math> i.e. <math>x = \tfrac{5}{3} - \tfrac{1}{3} = \tfrac{4}{3}\,\textrm{.}</math> | ||
+ | |||
+ | Check: | ||
+ | |||
+ | :*''x'' = 4/3: <math>\ \text{LHS} = 3\cdot\bigl(\tfrac{4}{3}\bigr)^{2} - 10\cdot\tfrac{4}{3} + 8 = 3\cdot\tfrac{16}{9} - \tfrac{40}{3} + \tfrac{8\cdot 3}{3} = 0 = \text{RHS,}</math> | ||
+ | |||
+ | :*''x'' = 2: <math>\ \text{LHS} = 3\cdot 2^{2} - 10\cdot 2 + 8 = 12 - 20 + 8 = 0 = \text{RHS.}</math> |
Current revision
We divide both sides by 3 and complete the square on the left-hand side,
\displaystyle \begin{align}
x^{2}-\frac{10}{3}x+\frac{8}{3} &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \Bigl(\frac{5}{3}\Bigr)^{2} + \frac{8}{3}\\[5pt] &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{25}{9} + \frac{24}{9}\\[5pt] &= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{1}{9}\,\textrm{.} \end{align} |
The equation then becomes
\displaystyle \left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,} |
and taking the square root gives the solutions as
- \displaystyle x-\tfrac{5}{3} = \sqrt{\tfrac{1}{9}} = \tfrac{1}{3}\,,\quad i.e. \displaystyle x = \tfrac{5}{3} + \tfrac{1}{3} = \tfrac{6}{3} = 2\,\textrm{,}
- \displaystyle x-\tfrac{5}{3} = -\sqrt{\tfrac{1}{9}} = -\tfrac{1}{3}\,,\quad i.e. \displaystyle x = \tfrac{5}{3} - \tfrac{1}{3} = \tfrac{4}{3}\,\textrm{.}
Check:
- x = 4/3: \displaystyle \ \text{LHS} = 3\cdot\bigl(\tfrac{4}{3}\bigr)^{2} - 10\cdot\tfrac{4}{3} + 8 = 3\cdot\tfrac{16}{9} - \tfrac{40}{3} + \tfrac{8\cdot 3}{3} = 0 = \text{RHS,}
- x = 2: \displaystyle \ \text{LHS} = 3\cdot 2^{2} - 10\cdot 2 + 8 = 12 - 20 + 8 = 0 = \text{RHS.}