Solution 3.2:5
From Förberedande kurs i matematik 1
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- | {{ | + | After squaring both sides, we obtain the equation |
- | < | + | |
- | {{ | + | {{Displayed math||<math>3x-2 = (2-x)^2</math>|(*)}} |
- | {{ | + | |
- | < | + | and if we expand the right-hand side and then collect the terms, we get |
- | {{ | + | |
+ | {{Displayed math||<math>x^{2}-7x+6=0\,\textrm{.}</math>}} | ||
+ | |||
+ | Completing the square of the left-hand side, we obtain | ||
+ | |||
+ | {{Displayed math||<math>\begin{align} | ||
+ | x^{2}-7x+6 | ||
+ | &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2+6\\[5pt] | ||
+ | &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{49}{4} + \frac{24}{4}\\[5pt] | ||
+ | &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{25}{4} | ||
+ | \end{align}</math>}} | ||
+ | |||
+ | which means that the equation can be written as | ||
+ | |||
+ | {{Displayed math||<math>\Bigl(x-\frac{7}{2}\Bigr)^2 = \frac{25}{4}</math>}} | ||
+ | |||
+ | and the solutions are therefore | ||
+ | |||
+ | :*<math>x = \frac{7}{2} + \sqrt{\frac{25}{4}} = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6\,,</math> | ||
+ | |||
+ | :*<math>x = \frac{7}{2} - \sqrt{\frac{25}{4}} = \frac{7}{2} - \frac{5}{2} = \frac{2}{2} = 1\,\textrm{.}</math> | ||
+ | |||
+ | Substituting <math>x=1</math> and <math>x=6</math> into the quadratic equation (*) shows that we have solved the equation correctly. | ||
+ | |||
+ | :*''x'' = 1: <math>\ \text{LHS} = 3\cdot 1-2 = 1\ </math> and <math>\ \text{RHS} = (2-1)^2 = 1</math> | ||
+ | |||
+ | :*''x'' = 6: <math>\ \text{LHS} = 3\cdot 6-2 = 16\ </math> and <math>\ \text{RHS} = (2-6)^2 = 16</math> | ||
+ | |||
+ | Finally, we need to sort away possible spurious roots to the root equation by verifying the solutions. | ||
+ | |||
+ | :*''x'' = 1: <math>\ \text{LHS} = \sqrt{3\cdot 1-2} = 1\ </math> and <math>\ \text{RHS} = 2-1 = 1</math> | ||
+ | |||
+ | :*''x'' = 6: <math>\ \text{LHS} = \sqrt{3\cdot 6-2} = 4\ </math> and <math>\ \text{RHS} = 2-6 = -4</math> | ||
+ | |||
+ | This shows that the root equation has the solution <math>x=1\,</math>. |
Current revision
After squaring both sides, we obtain the equation
\displaystyle 3x-2 = (2-x)^2 | (*) |
and if we expand the right-hand side and then collect the terms, we get
\displaystyle x^{2}-7x+6=0\,\textrm{.} |
Completing the square of the left-hand side, we obtain
\displaystyle \begin{align}
x^{2}-7x+6 &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2+6\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{49}{4} + \frac{24}{4}\\[5pt] &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{25}{4} \end{align} |
which means that the equation can be written as
\displaystyle \Bigl(x-\frac{7}{2}\Bigr)^2 = \frac{25}{4} |
and the solutions are therefore
- \displaystyle x = \frac{7}{2} + \sqrt{\frac{25}{4}} = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6\,,
- \displaystyle x = \frac{7}{2} - \sqrt{\frac{25}{4}} = \frac{7}{2} - \frac{5}{2} = \frac{2}{2} = 1\,\textrm{.}
Substituting \displaystyle x=1 and \displaystyle x=6 into the quadratic equation (*) shows that we have solved the equation correctly.
- x = 1: \displaystyle \ \text{LHS} = 3\cdot 1-2 = 1\ and \displaystyle \ \text{RHS} = (2-1)^2 = 1
- x = 6: \displaystyle \ \text{LHS} = 3\cdot 6-2 = 16\ and \displaystyle \ \text{RHS} = (2-6)^2 = 16
Finally, we need to sort away possible spurious roots to the root equation by verifying the solutions.
- x = 1: \displaystyle \ \text{LHS} = \sqrt{3\cdot 1-2} = 1\ and \displaystyle \ \text{RHS} = 2-1 = 1
- x = 6: \displaystyle \ \text{LHS} = \sqrt{3\cdot 6-2} = 4\ and \displaystyle \ \text{RHS} = 2-6 = -4
This shows that the root equation has the solution \displaystyle x=1\,.