Solution 2.3:2e
From Förberedande kurs i matematik 1
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- | {{ | + | Write the equation in normalized form by dividing both sides by 5, |
- | < | + | |
- | {{ | + | {{Displayed math||<math>x^{2}+\frac{2}{5}x-\frac{3}{5}=0\,\textrm{.}</math>}} |
+ | |||
+ | Complete the square on the left-hand side, | ||
+ | |||
+ | {{Displayed math||<math>\begin{align} | ||
+ | x^{2}+\frac{2}{5}x-\frac{3}{5} | ||
+ | &= \Bigl(x+\frac{2/5}{2}\Bigr)^{2} - \Bigl(\frac{2/5}{2}\Bigr)^{2} - \frac{3}{5}\\[5pt] | ||
+ | &= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \Bigl(\frac{1}{5}\Bigr)^{2} - \frac{3}{5}\\[5pt] | ||
+ | &= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \frac{1}{25} - \frac{3\cdot 5}{25}\\[5pt] | ||
+ | &= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \frac{16}{25}\,\textrm{.} | ||
+ | \end{align}</math>}} | ||
+ | |||
+ | The equation is now rewritten as | ||
+ | |||
+ | {{Displayed math||<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}\,\textrm{,}</math>}} | ||
+ | |||
+ | and taking the root gives the solutions | ||
+ | |||
+ | :*<math>x+\tfrac{1}{5} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5}</math> because <math>\bigl(\tfrac{4}{5}\bigr)^{2} = \tfrac{16}{25}\,,</math> which gives <math>x=-\tfrac{1}{5}+\tfrac{4}{5}=\tfrac{3}{5},</math> | ||
+ | |||
+ | :*<math>x+\tfrac{1}{5} = -\sqrt{\tfrac{16}{25}} = -\tfrac{4}{5}\,,</math> which gives <math>x = -\tfrac{1}{5}-\tfrac{4}{5}=-1\,\textrm{.}</math> | ||
+ | |||
+ | Finally, we check the answer by substituting <math>x=-1</math> and <math>x=3/5</math> into the equation: | ||
+ | |||
+ | :*''x'' = 1: <math>\ \text{LHS} = 5\cdot (-1)^{2} + 2\cdot (-1) - 3 = 5 - 2 - 3 = 0 = \text{RHS,}</math> | ||
+ | |||
+ | :*''x'' = 3/5: <math>\ \text{LHS} = 5\cdot\bigl(\tfrac{3}{5}\bigr)^{2} + 2\cdot\bigl(\tfrac{3}{5}\bigr) - 3 = 5\cdot\tfrac{9}{25} + \tfrac{6}{5} - \tfrac{3\cdot 5}{5} = 0 = \text{RHS.}</math> |
Current revision
Write the equation in normalized form by dividing both sides by 5,
\displaystyle x^{2}+\frac{2}{5}x-\frac{3}{5}=0\,\textrm{.} |
Complete the square on the left-hand side,
\displaystyle \begin{align}
x^{2}+\frac{2}{5}x-\frac{3}{5} &= \Bigl(x+\frac{2/5}{2}\Bigr)^{2} - \Bigl(\frac{2/5}{2}\Bigr)^{2} - \frac{3}{5}\\[5pt] &= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \Bigl(\frac{1}{5}\Bigr)^{2} - \frac{3}{5}\\[5pt] &= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \frac{1}{25} - \frac{3\cdot 5}{25}\\[5pt] &= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \frac{16}{25}\,\textrm{.} \end{align} |
The equation is now rewritten as
\displaystyle \left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}\,\textrm{,} |
and taking the root gives the solutions
- \displaystyle x+\tfrac{1}{5} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5} because \displaystyle \bigl(\tfrac{4}{5}\bigr)^{2} = \tfrac{16}{25}\,, which gives \displaystyle x=-\tfrac{1}{5}+\tfrac{4}{5}=\tfrac{3}{5},
- \displaystyle x+\tfrac{1}{5} = -\sqrt{\tfrac{16}{25}} = -\tfrac{4}{5}\,, which gives \displaystyle x = -\tfrac{1}{5}-\tfrac{4}{5}=-1\,\textrm{.}
Finally, we check the answer by substituting \displaystyle x=-1 and \displaystyle x=3/5 into the equation:
- x = 1: \displaystyle \ \text{LHS} = 5\cdot (-1)^{2} + 2\cdot (-1) - 3 = 5 - 2 - 3 = 0 = \text{RHS,}
- x = 3/5: \displaystyle \ \text{LHS} = 5\cdot\bigl(\tfrac{3}{5}\bigr)^{2} + 2\cdot\bigl(\tfrac{3}{5}\bigr) - 3 = 5\cdot\tfrac{9}{25} + \tfrac{6}{5} - \tfrac{3\cdot 5}{5} = 0 = \text{RHS.}