If √(x + √5) − √(x − √5) = √2, then the value of x is:
Answer and solution
Correct answer: 3
Solution
Answer: 3
1. Start by squaring both sides:
2. [√(x + √5) − √(x − √5)]² = 2
3. After expanding:
4. x + √5 + x − √5 − 2√(x² − 5) = 2
5. So,
6. 2x − 2√(x² − 5) = 2
x − √(x² − 5) = 1
7. Therefore,
8. √(x² − 5) = x − 1
9. Squaring again:
10. x² − 5 = (x − 1)²
x² − 5 = x² − 2x + 1
2x = 6
x = 3
11. Answer: 3 (TITA).
12. Domain and substitution: x must be at least √5. At x=3, both radicals are real and their difference is positive with square 2, so the original equation holds.
Why this method works:
The domain x≥√5 ensures both original square roots exist. Squaring the difference introduces a cross term −2√(x²−5); omitting this term is the main algebra trap. Isolate the remaining radical before squaring a second time.
Independent check and common trap:
At x=3 the larger radical exceeds the smaller, so their difference is positive. Its square is 6−2√4=2, proving that the difference equals +√2 rather than −√2. This validates the candidate in the unsquared equation.
Apply: Square once: x−√(x²−5)=1. Square the isolated radical: 2x=6 →x=3. Substitute to verify the positive square-root difference.
Check: At x=3 the larger radical exceeds the smaller, so their difference is positive. Its square is 6−2√4=2, proving that the difference equals +√2 rather than −√2. This validates the candidate in the unsquared equation.
1. Start by squaring both sides:
2. [√(x + √5) − √(x − √5)]² = 2
3. After expanding:
4. x + √5 + x − √5 − 2√(x² − 5) = 2
5. So,
6. 2x − 2√(x² − 5) = 2
x − √(x² − 5) = 1
7. Therefore,
8. √(x² − 5) = x − 1
9. Squaring again:
10. x² − 5 = (x − 1)²
x² − 5 = x² − 2x + 1
2x = 6
x = 3
11. Answer: 3 (TITA).
12. Domain and substitution: x must be at least √5. At x=3, both radicals are real and their difference is positive with square 2, so the original equation holds.
Why this method works:
The domain x≥√5 ensures both original square roots exist. Squaring the difference introduces a cross term −2√(x²−5); omitting this term is the main algebra trap. Isolate the remaining radical before squaring a second time.
Independent check and common trap:
At x=3 the larger radical exceeds the smaller, so their difference is positive. Its square is 6−2√4=2, proving that the difference equals +√2 rather than −√2. This validates the candidate in the unsquared equation.
G Strategy shortcut
Spot: Algebra – Surds and Equations.Apply: Square once: x−√(x²−5)=1. Square the isolated radical: 2x=6 →x=3. Substitute to verify the positive square-root difference.
Check: At x=3 the larger radical exceeds the smaller, so their difference is positive. Its square is 6−2√4=2, proving that the difference equals +√2 rather than −√2. This validates the candidate in the unsquared equation.
G Strategy: Spot: Algebra – Surds and Equations. Apply: Square once: x−√(x²−5)=1. Square the isolated radical: 2x=6 →x=3. Substitute to verify the positive square-root difference. Check: At x=3 the larger radical exceeds the smaller, so their difference is positive. Its square is 6−2√4=2, proving that the difference equals +√2 rather than −√2. This validates the candidate in the unsquared equation.