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Sagot :
To solve the equation [tex]\(\sqrt{3x + 16} = x + 2\)[/tex], let's go through the steps methodically.
1. Square both sides to eliminate the square root:
[tex]\[ (\sqrt{3x + 16})^2 = (x + 2)^2 \][/tex]
This gives:
[tex]\[ 3x + 16 = x^2 + 4x + 4 \][/tex]
2. Rearrange the equation to set it to zero:
[tex]\[ x^2 + 4x + 4 - 3x - 16 = 0 \][/tex]
Simplify the terms:
[tex]\[ x^2 + x - 12 = 0 \][/tex]
3. Factor the quadratic equation:
[tex]\[ x^2 + x - 12 = (x - 3)(x + 4) = 0 \][/tex]
4. Find the roots:
[tex]\[ x - 3 = 0 \quad \text{or} \quad x + 4 = 0 \][/tex]
Thus,
[tex]\[ x = 3 \quad \text{or} \quad x = -4 \][/tex]
5. Verify the roots by substituting them back into the original equation to check for extraneous solutions:
- For [tex]\(x = 3\)[/tex]:
[tex]\[ \sqrt{3(3) + 16} = 3 + 2 \][/tex]
[tex]\[ \sqrt{9 + 16} = 5 \][/tex]
[tex]\[ \sqrt{25} = 5 \][/tex]
This is true, so [tex]\(x = 3\)[/tex] is a valid solution.
- For [tex]\(x = -4\)[/tex]:
[tex]\[ \sqrt{3(-4) + 16} = -4 + 2 \][/tex]
[tex]\[ \sqrt{-12 + 16} = -2 \][/tex]
[tex]\[ \sqrt{4} = -2 \][/tex]
Since [tex]\(\sqrt{4} = 2\)[/tex] and not [tex]\(-2\)[/tex], [tex]\(x = -4\)[/tex] is not a valid solution.
After checking, we find that the only valid solution is:
[tex]\[ \{3\} \][/tex]
Therefore, the correct answer is:
3) [tex]\(\{3\}\)[/tex]
1. Square both sides to eliminate the square root:
[tex]\[ (\sqrt{3x + 16})^2 = (x + 2)^2 \][/tex]
This gives:
[tex]\[ 3x + 16 = x^2 + 4x + 4 \][/tex]
2. Rearrange the equation to set it to zero:
[tex]\[ x^2 + 4x + 4 - 3x - 16 = 0 \][/tex]
Simplify the terms:
[tex]\[ x^2 + x - 12 = 0 \][/tex]
3. Factor the quadratic equation:
[tex]\[ x^2 + x - 12 = (x - 3)(x + 4) = 0 \][/tex]
4. Find the roots:
[tex]\[ x - 3 = 0 \quad \text{or} \quad x + 4 = 0 \][/tex]
Thus,
[tex]\[ x = 3 \quad \text{or} \quad x = -4 \][/tex]
5. Verify the roots by substituting them back into the original equation to check for extraneous solutions:
- For [tex]\(x = 3\)[/tex]:
[tex]\[ \sqrt{3(3) + 16} = 3 + 2 \][/tex]
[tex]\[ \sqrt{9 + 16} = 5 \][/tex]
[tex]\[ \sqrt{25} = 5 \][/tex]
This is true, so [tex]\(x = 3\)[/tex] is a valid solution.
- For [tex]\(x = -4\)[/tex]:
[tex]\[ \sqrt{3(-4) + 16} = -4 + 2 \][/tex]
[tex]\[ \sqrt{-12 + 16} = -2 \][/tex]
[tex]\[ \sqrt{4} = -2 \][/tex]
Since [tex]\(\sqrt{4} = 2\)[/tex] and not [tex]\(-2\)[/tex], [tex]\(x = -4\)[/tex] is not a valid solution.
After checking, we find that the only valid solution is:
[tex]\[ \{3\} \][/tex]
Therefore, the correct answer is:
3) [tex]\(\{3\}\)[/tex]
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