Solution 2.3:2b
From Förberedande kurs i matematik 1
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- | {{ | + | The first step when we solve the second-degree equation is to complete the square on the left-hand side |
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- | {{ | + | {{Displayed math||<math>y^{2}+2y-15 = (y+1)^{2}-1^{2}-15 = (y+1)^{2}-16\,\textrm{.}</math>}} |
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+ | The equation can now be written as | ||
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+ | {{Displayed math||<math>(y+1)^{2} = 16</math>}} | ||
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+ | and has, after taking the square root, the solutions: | ||
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+ | :*<math>y+1 = \sqrt{16} = 4\,\textrm{,}\ </math> which gives <math>y=-1+4=3\,\textrm{,}</math> | ||
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+ | :*<math>y+1 = -\sqrt{16} = -4\,\textrm{,}\ </math> which gives <math>y=-1-4=-5\,\textrm{.}</math> | ||
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+ | A quick check shows that <math>y=-5</math> and <math>y=3</math> satisfy the equation: | ||
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+ | :*''y'' = -5: <math>\ \text{LHS} = (-5)^{2} + 2\cdot (-5)-15 = 25-10-15 = 0 = \text{RHS,}</math> | ||
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+ | :*''y'' = 3: <math>\ \text{LHS} = 3^{2} + 2\cdot 3 - 15 = 9+6-15 = 0 = \text{RHS.}</math> |
Current revision
The first step when we solve the second-degree equation is to complete the square on the left-hand side
\displaystyle y^{2}+2y-15 = (y+1)^{2}-1^{2}-15 = (y+1)^{2}-16\,\textrm{.} |
The equation can now be written as
\displaystyle (y+1)^{2} = 16 |
and has, after taking the square root, the solutions:
- \displaystyle y+1 = \sqrt{16} = 4\,\textrm{,}\ which gives \displaystyle y=-1+4=3\,\textrm{,}
- \displaystyle y+1 = -\sqrt{16} = -4\,\textrm{,}\ which gives \displaystyle y=-1-4=-5\,\textrm{.}
A quick check shows that \displaystyle y=-5 and \displaystyle y=3 satisfy the equation:
- y = -5: \displaystyle \ \text{LHS} = (-5)^{2} + 2\cdot (-5)-15 = 25-10-15 = 0 = \text{RHS,}
- y = 3: \displaystyle \ \text{LHS} = 3^{2} + 2\cdot 3 - 15 = 9+6-15 = 0 = \text{RHS.}