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Sagot :
To determine which equation can be rewritten as [tex]\( x + 4 = x^2 \)[/tex] with the condition [tex]\( x > 0 \)[/tex], we will analyze each given equation step by step.
### Equation 1: [tex]\(\sqrt{x} + 2 = x\)[/tex]
1. Start with [tex]\(\sqrt{x} + 2 = x\)[/tex].
2. Isolate the square root term: [tex]\(\sqrt{x} = x - 2\)[/tex].
3. Square both sides: [tex]\(x = (x - 2)^2\)[/tex].
4. Expand the right side: [tex]\(x = x^2 - 4x + 4\)[/tex].
5. Rearrange the equation: [tex]\(x = x^2 - 4x + 4\)[/tex].
6. Collect all terms to one side: [tex]\(x^2 - 5x + 4 = 0\)[/tex].
7. Factor the quadratic: [tex]\((x - 1)(x - 4) = 0\)[/tex].
8. Solve for [tex]\(x\)[/tex]: [tex]\(x = 1\)[/tex] or [tex]\(x = 4\)[/tex].
With [tex]\( x > 0 \)[/tex], both solutions [tex]\( x = 1 \)[/tex] and [tex]\( x = 4 \)[/tex] are valid solutions, but we will determine if the equation fits [tex]\( x + 4 = x^2 \)[/tex] later.
### Equation 2: [tex]\(\sqrt{x + 2} = x\)[/tex]
1. Start with [tex]\(\sqrt{x + 2} = x\)[/tex].
2. Square both sides: [tex]\(x + 2 = x^2\)[/tex].
3. Rearrange the equation: [tex]\(x^2 - x - 2 = 0\)[/tex].
4. Factor the quadratic: [tex]\((x - 2)(x + 1) = 0\)[/tex].
5. Solve for [tex]\(x\)[/tex]: [tex]\(x = 2\)[/tex] or [tex]\(x = -1\)[/tex].
With [tex]\( x > 0 \)[/tex], the valid solution is [tex]\( x = 2 \)[/tex].
Check how it fits with [tex]\(x + 4 = x^2\)[/tex]:
- For [tex]\( x = 2 \)[/tex]: [tex]\( 2 + 4 = 6 \)[/tex] and [tex]\( 2^2 = 4 \)[/tex]. It does not fit.
### Equation 3: [tex]\(\sqrt{x + 4} = x\)[/tex]
1. Start with [tex]\(\sqrt{x + 4} = x\)[/tex].
2. Square both sides: [tex]\(x + 4 = x^2\)[/tex].
3. Rearrange: [tex]\(x^2 - x - 4 = 0\)[/tex].
This is the form [tex]\(x + 4 = x^2\)[/tex]!
Solve it to check potential [tex]\(x\)[/tex]:
- Factorization might be tedious so quadratic solution:
[tex]\[x = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2}\][/tex]
With [tex]\( x > 0 \)[/tex], only positive [tex]\(x = \frac{1 + \sqrt{17}}{2}\)[/tex].
### Equation 4: [tex]\(\sqrt{x^2 + 16} = x\)[/tex]
1. Start with [tex]\(\sqrt{x^2 + 16} = x\)[/tex].
2. Square both sides: [tex]\(x^2 + 16 = x^2\)[/tex].
3. Simplify: [tex]\(16 = 0\)[/tex].
This provides no valid [tex]\(x\)[/tex].
### Conclusion
From the analysis, the equation that can be rewritten as [tex]\( x + 4 = x^2 \)[/tex] is:
[tex]\[ \sqrt{x + 4} = x \][/tex]
### Equation 1: [tex]\(\sqrt{x} + 2 = x\)[/tex]
1. Start with [tex]\(\sqrt{x} + 2 = x\)[/tex].
2. Isolate the square root term: [tex]\(\sqrt{x} = x - 2\)[/tex].
3. Square both sides: [tex]\(x = (x - 2)^2\)[/tex].
4. Expand the right side: [tex]\(x = x^2 - 4x + 4\)[/tex].
5. Rearrange the equation: [tex]\(x = x^2 - 4x + 4\)[/tex].
6. Collect all terms to one side: [tex]\(x^2 - 5x + 4 = 0\)[/tex].
7. Factor the quadratic: [tex]\((x - 1)(x - 4) = 0\)[/tex].
8. Solve for [tex]\(x\)[/tex]: [tex]\(x = 1\)[/tex] or [tex]\(x = 4\)[/tex].
With [tex]\( x > 0 \)[/tex], both solutions [tex]\( x = 1 \)[/tex] and [tex]\( x = 4 \)[/tex] are valid solutions, but we will determine if the equation fits [tex]\( x + 4 = x^2 \)[/tex] later.
### Equation 2: [tex]\(\sqrt{x + 2} = x\)[/tex]
1. Start with [tex]\(\sqrt{x + 2} = x\)[/tex].
2. Square both sides: [tex]\(x + 2 = x^2\)[/tex].
3. Rearrange the equation: [tex]\(x^2 - x - 2 = 0\)[/tex].
4. Factor the quadratic: [tex]\((x - 2)(x + 1) = 0\)[/tex].
5. Solve for [tex]\(x\)[/tex]: [tex]\(x = 2\)[/tex] or [tex]\(x = -1\)[/tex].
With [tex]\( x > 0 \)[/tex], the valid solution is [tex]\( x = 2 \)[/tex].
Check how it fits with [tex]\(x + 4 = x^2\)[/tex]:
- For [tex]\( x = 2 \)[/tex]: [tex]\( 2 + 4 = 6 \)[/tex] and [tex]\( 2^2 = 4 \)[/tex]. It does not fit.
### Equation 3: [tex]\(\sqrt{x + 4} = x\)[/tex]
1. Start with [tex]\(\sqrt{x + 4} = x\)[/tex].
2. Square both sides: [tex]\(x + 4 = x^2\)[/tex].
3. Rearrange: [tex]\(x^2 - x - 4 = 0\)[/tex].
This is the form [tex]\(x + 4 = x^2\)[/tex]!
Solve it to check potential [tex]\(x\)[/tex]:
- Factorization might be tedious so quadratic solution:
[tex]\[x = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2}\][/tex]
With [tex]\( x > 0 \)[/tex], only positive [tex]\(x = \frac{1 + \sqrt{17}}{2}\)[/tex].
### Equation 4: [tex]\(\sqrt{x^2 + 16} = x\)[/tex]
1. Start with [tex]\(\sqrt{x^2 + 16} = x\)[/tex].
2. Square both sides: [tex]\(x^2 + 16 = x^2\)[/tex].
3. Simplify: [tex]\(16 = 0\)[/tex].
This provides no valid [tex]\(x\)[/tex].
### Conclusion
From the analysis, the equation that can be rewritten as [tex]\( x + 4 = x^2 \)[/tex] is:
[tex]\[ \sqrt{x + 4} = x \][/tex]
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