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Sagot :
To solve the equation [tex]\( x - 2 = \sqrt{x - 2} + 12 \)[/tex], we will proceed through the following steps:
1. Isolate the square root term:
[tex]\[ x - 2 - 12 = \sqrt{x - 2} \][/tex]
Simplifying the left-hand side, we get:
[tex]\[ x - 14 = \sqrt{x - 2} \][/tex]
2. Square both sides to eliminate the square root:
[tex]\[ (x - 14)^2 = (\sqrt{x - 2})^2 \][/tex]
Simplifying both sides, we get:
[tex]\[ (x - 14)^2 = x - 2 \][/tex]
3. Expand the left-hand side:
[tex]\[ x^2 - 28x + 196 = x - 2 \][/tex]
4. Move all terms to one side to form a quadratic equation:
[tex]\[ x^2 - 28x + 196 - x + 2 = 0 \][/tex]
Simplifying the equation:
[tex]\[ x^2 - 29x + 198 = 0 \][/tex]
5. Solve the quadratic equation:
We use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -29 \)[/tex], and [tex]\( c = 198 \)[/tex].
Computing the discriminant:
[tex]\[ b^2 - 4ac = (-29)^2 - 4 \cdot 1 \cdot 198 = 841 - 792 = 49 \][/tex]
Thus:
[tex]\[ x = \frac{29 \pm \sqrt{49}}{2 \cdot 1} = \frac{29 \pm 7}{2} \][/tex]
This gives us two potential solutions:
[tex]\[ x = \frac{29 + 7}{2} = \frac{36}{2} = 18 \][/tex]
[tex]\[ x = \frac{29 - 7}{2} = \frac{22}{2} = 11 \][/tex]
6. Check each potential solution:
- For [tex]\( x = 18 \)[/tex]:
[tex]\[ 18 - 2 = \sqrt{18 - 2} + 12 \][/tex]
Simplifying,
[tex]\[ 16 = \sqrt{16} + 12 \][/tex]
[tex]\[ 16 = 4 + 12 \][/tex]
[tex]\[ 16 = 16 \quad (\text{True}) \][/tex]
So, [tex]\( x = 18 \)[/tex] is a valid solution.
- For [tex]\( x = 11 \)[/tex]:
[tex]\[ 11 - 2 = \sqrt{11 - 2} + 12 \][/tex]
Simplifying,
[tex]\[ 9 = \sqrt{9} + 12 \][/tex]
[tex]\[ 9 = 3 + 12 \][/tex]
[tex]\[ 9 \neq 15 \quad (\text{False}) \][/tex]
So, [tex]\( x = 11 \)[/tex] is not a valid solution.
Therefore, the solution to the equation [tex]\( x - 2 = \sqrt{x - 2} + 12 \)[/tex] is:
[tex]\[ \boxed{18} \][/tex]
1. Isolate the square root term:
[tex]\[ x - 2 - 12 = \sqrt{x - 2} \][/tex]
Simplifying the left-hand side, we get:
[tex]\[ x - 14 = \sqrt{x - 2} \][/tex]
2. Square both sides to eliminate the square root:
[tex]\[ (x - 14)^2 = (\sqrt{x - 2})^2 \][/tex]
Simplifying both sides, we get:
[tex]\[ (x - 14)^2 = x - 2 \][/tex]
3. Expand the left-hand side:
[tex]\[ x^2 - 28x + 196 = x - 2 \][/tex]
4. Move all terms to one side to form a quadratic equation:
[tex]\[ x^2 - 28x + 196 - x + 2 = 0 \][/tex]
Simplifying the equation:
[tex]\[ x^2 - 29x + 198 = 0 \][/tex]
5. Solve the quadratic equation:
We use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -29 \)[/tex], and [tex]\( c = 198 \)[/tex].
Computing the discriminant:
[tex]\[ b^2 - 4ac = (-29)^2 - 4 \cdot 1 \cdot 198 = 841 - 792 = 49 \][/tex]
Thus:
[tex]\[ x = \frac{29 \pm \sqrt{49}}{2 \cdot 1} = \frac{29 \pm 7}{2} \][/tex]
This gives us two potential solutions:
[tex]\[ x = \frac{29 + 7}{2} = \frac{36}{2} = 18 \][/tex]
[tex]\[ x = \frac{29 - 7}{2} = \frac{22}{2} = 11 \][/tex]
6. Check each potential solution:
- For [tex]\( x = 18 \)[/tex]:
[tex]\[ 18 - 2 = \sqrt{18 - 2} + 12 \][/tex]
Simplifying,
[tex]\[ 16 = \sqrt{16} + 12 \][/tex]
[tex]\[ 16 = 4 + 12 \][/tex]
[tex]\[ 16 = 16 \quad (\text{True}) \][/tex]
So, [tex]\( x = 18 \)[/tex] is a valid solution.
- For [tex]\( x = 11 \)[/tex]:
[tex]\[ 11 - 2 = \sqrt{11 - 2} + 12 \][/tex]
Simplifying,
[tex]\[ 9 = \sqrt{9} + 12 \][/tex]
[tex]\[ 9 = 3 + 12 \][/tex]
[tex]\[ 9 \neq 15 \quad (\text{False}) \][/tex]
So, [tex]\( x = 11 \)[/tex] is not a valid solution.
Therefore, the solution to the equation [tex]\( x - 2 = \sqrt{x - 2} + 12 \)[/tex] is:
[tex]\[ \boxed{18} \][/tex]
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