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Sagot :
To solve the equation [tex]\(\sqrt{x + 9} - 4 = 1\)[/tex] for [tex]\(x\)[/tex], follow these detailed steps:
1. Isolate the Square Root Term:
[tex]\[ \sqrt{x + 9} - 4 = 1 \][/tex]
Add 4 to both sides of the equation to isolate the square root term:
[tex]\[ \sqrt{x + 9} = 1 + 4 \][/tex]
Simplify the right-hand side:
[tex]\[ \sqrt{x + 9} = 5 \][/tex]
2. Square Both Sides to Eliminate the Square Root:
[tex]\[ (\sqrt{x + 9})^2 = 5^2 \][/tex]
This reduces to:
[tex]\[ x + 9 = 25 \][/tex]
3. Solve for [tex]\(x\)[/tex]:
Subtract 9 from both sides of the equation to solve for [tex]\(x\)[/tex]:
[tex]\[ x = 25 - 9 \][/tex]
Simplify the right-hand side:
[tex]\[ x = 16 \][/tex]
4. Check for Extraneous Solutions:
Substitute [tex]\(x = 16\)[/tex] back into the original equation to ensure it satisfies the equation:
[tex]\[ \sqrt{16 + 9} - 4 = 1 \][/tex]
Simplify inside the square root:
[tex]\[ \sqrt{25} - 4 = 1 \][/tex]
Since [tex]\(\sqrt{25} = 5\)[/tex], we have:
[tex]\[ 5 - 4 = 1 \][/tex]
This simplifies correctly to:
[tex]\[ 1 = 1 \][/tex]
The solution [tex]\(x = 16\)[/tex] satisfies the original equation, so it is not extraneous.
Therefore, the correct answer is:
[tex]\[ \boxed{x = 16, \text{ solution is not extraneous}} \][/tex]
1. Isolate the Square Root Term:
[tex]\[ \sqrt{x + 9} - 4 = 1 \][/tex]
Add 4 to both sides of the equation to isolate the square root term:
[tex]\[ \sqrt{x + 9} = 1 + 4 \][/tex]
Simplify the right-hand side:
[tex]\[ \sqrt{x + 9} = 5 \][/tex]
2. Square Both Sides to Eliminate the Square Root:
[tex]\[ (\sqrt{x + 9})^2 = 5^2 \][/tex]
This reduces to:
[tex]\[ x + 9 = 25 \][/tex]
3. Solve for [tex]\(x\)[/tex]:
Subtract 9 from both sides of the equation to solve for [tex]\(x\)[/tex]:
[tex]\[ x = 25 - 9 \][/tex]
Simplify the right-hand side:
[tex]\[ x = 16 \][/tex]
4. Check for Extraneous Solutions:
Substitute [tex]\(x = 16\)[/tex] back into the original equation to ensure it satisfies the equation:
[tex]\[ \sqrt{16 + 9} - 4 = 1 \][/tex]
Simplify inside the square root:
[tex]\[ \sqrt{25} - 4 = 1 \][/tex]
Since [tex]\(\sqrt{25} = 5\)[/tex], we have:
[tex]\[ 5 - 4 = 1 \][/tex]
This simplifies correctly to:
[tex]\[ 1 = 1 \][/tex]
The solution [tex]\(x = 16\)[/tex] satisfies the original equation, so it is not extraneous.
Therefore, the correct answer is:
[tex]\[ \boxed{x = 16, \text{ solution is not extraneous}} \][/tex]
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