Solution 3.2:2
From Förberedande kurs i matematik 1
(Difference between revisions)
m |
m |
||
| Line 1: | Line 1: | ||
The first thing we do is to square both sides of the equation | The first thing we do is to square both sides of the equation | ||
| + | {{Displayed math||<math>2x+7 = (x+2)^2</math>}} | ||
| - | + | to obtain an equation without a root sign. It is possible that we thereby introduce so-called spurious roots (solutions to the new equation which are not solutions to the old equation), so we need to test the solutions in the original root equation before we answer. | |
| - | + | ||
| - | + | ||
| - | to | + | |
If we expand the right-hand side in the squared equation, we get | If we expand the right-hand side in the squared equation, we get | ||
| - | + | {{Displayed math||<math>2x+7=x^{2}+4x+4</math>|(*)}} | |
| - | + | ||
| - | <math>2x+7=x^{2}+4x+4</math> | + | |
| - | + | ||
which we also can write as | which we also can write as | ||
| - | + | {{Displayed math||<math>x^{2}+2x-3=0\,\textrm{.}</math>}} | |
| - | <math>x^{2}+2x-3=0</math> | + | |
| - | + | ||
Completing the square of the left-hand side gives | Completing the square of the left-hand side gives | ||
| + | {{Displayed math||<math>x^2+2x-3 = (x+1)^2-1^2-3 = (x+1)^2-4\,\textrm{.}</math>}} | ||
| - | <math>x^{2}+2x-3=\left( x+1 \right)^{2}-1^{2}-3=\left( x+1 \right)^{2}-4.</math> | ||
| - | |||
| - | |||
The equation then becomes | The equation then becomes | ||
| - | + | {{Displayed math||<math>(x+1)^2 = 4</math>}} | |
| - | <math> | + | |
| - | + | ||
which has solutions | which has solutions | ||
| + | :*<math>x=-1+\sqrt{4}=-1+2=1</math> | ||
| - | + | :*<math>x=-1-\sqrt{4}=-1-2=-3</math> | |
| - | + | ||
| - | + | ||
| - | <math>x=-1-\sqrt{4}=-1-2=-3 | + | |
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | A quick check shows also that <math>x=-3</math> and <math>x=1</math> are solutions to the squared equation (*): | |
| - | <math>=-3 | + | |
| + | :{| | ||
| + | ||<ul><li>''x'' = -3:</li></ul> | ||
| + | ||<math>\ \text{LHS} = 2\cdot (-3)+7 = -6+7 = 1</math> and | ||
| + | |- | ||
| + | || | ||
| + | ||<math>\ \text{RHS} = (-3+2)^{2} = 1</math> | ||
| + | |- | ||
| + | ||<ul><li>''x'' = 1:</li></ul> | ||
| + | ||<math>\ \text{LHS} = 2\cdot 1+7 = 2+7 = 9</math> and | ||
| + | |- | ||
| + | || | ||
| + | ||<math>\ \text{RHS} = (1+2)^2 = 9</math> | ||
| + | |} | ||
| - | + | When we test the solutions in the root equation, we get that | |
| - | + | ||
| - | RHS | + | :{| |
| - | <math>=1+2=3</math> | + | ||<ul><li>''x'' = -3:</li></ul> |
| + | ||<math>\ \text{LHS} = \sqrt{2\cdot (-3)+7} = \sqrt{-6+7} = \sqrt{1} = 1</math> and | ||
| + | |- | ||
| + | || | ||
| + | ||<math>\ \text{RHS} = -3+2 = -1</math> | ||
| + | |- | ||
| + | ||<ul><li>''x'' = 1:</li></ul> | ||
| + | ||<math>\ \text{LHS} = \sqrt{2\cdot 1+7} = \sqrt{2+7} = \sqrt{9} = 3</math> | ||
| + | |- | ||
| + | || | ||
| + | ||<math>\ \text{RHS} = 1+2 = 3</math> | ||
| + | |} | ||
| + | and therefore <math>x=1</math> is the only solution to the root equation (<math>x=-3</math> is a spurious root). | ||
| - | and therefore | ||
| - | <math>x=\text{1}</math> | ||
| - | is the only solution to the root equation ( | ||
| - | <math>x=-\text{3}</math> | ||
| - | is a false root). | ||
| - | + | Note: The check we carry out when substituting the solutions into equation (*) is not strictly speaking necessary, but more for seeing that we haven't calculated incorrectly. On the other hand, testing in the root equation is necessary. | |
Current revision
The first thing we do is to square both sides of the equation
| \displaystyle 2x+7 = (x+2)^2 |
to obtain an equation without a root sign. It is possible that we thereby introduce so-called spurious roots (solutions to the new equation which are not solutions to the old equation), so we need to test the solutions in the original root equation before we answer.
If we expand the right-hand side in the squared equation, we get
| \displaystyle 2x+7=x^{2}+4x+4 | (*) |
which we also can write as
| \displaystyle x^{2}+2x-3=0\,\textrm{.} |
Completing the square of the left-hand side gives
| \displaystyle x^2+2x-3 = (x+1)^2-1^2-3 = (x+1)^2-4\,\textrm{.} |
The equation then becomes
| \displaystyle (x+1)^2 = 4 |
which has solutions
- \displaystyle x=-1+\sqrt{4}=-1+2=1
- \displaystyle x=-1-\sqrt{4}=-1-2=-3
A quick check shows also that \displaystyle x=-3 and \displaystyle x=1 are solutions to the squared equation (*):
- x = -3:
\displaystyle \ \text{LHS} = 2\cdot (-3)+7 = -6+7 = 1 and \displaystyle \ \text{RHS} = (-3+2)^{2} = 1 - x = 1:
\displaystyle \ \text{LHS} = 2\cdot 1+7 = 2+7 = 9 and \displaystyle \ \text{RHS} = (1+2)^2 = 9
When we test the solutions in the root equation, we get that
- x = -3:
\displaystyle \ \text{LHS} = \sqrt{2\cdot (-3)+7} = \sqrt{-6+7} = \sqrt{1} = 1 and \displaystyle \ \text{RHS} = -3+2 = -1 - x = 1:
\displaystyle \ \text{LHS} = \sqrt{2\cdot 1+7} = \sqrt{2+7} = \sqrt{9} = 3 \displaystyle \ \text{RHS} = 1+2 = 3
and therefore \displaystyle x=1 is the only solution to the root equation (\displaystyle x=-3 is a spurious root).
Note: The check we carry out when substituting the solutions into equation (*) is not strictly speaking necessary, but more for seeing that we haven't calculated incorrectly. On the other hand, testing in the root equation is necessary.
