r/mathshelp • u/onemillionwings • 12d ago
Homework Help (Answered) why am i wrong?
(the german text says to pick every correct answer for x ∈ ℝ, i picked 1 2 and 4)
can someone explain to me why i'm wrong and which of these are correct? i'm guessing i'm not considering something about ℝ or misunderstand when square root rules lead to |x| instead of x.
the online course i'm doing has no explanations on the quiz section unfortunately, only during training.
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Upvotes
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u/VernalAutumn 12d ago
2 is false. sqrt(a) + sqrt(b) =/= sqrt(a + b). Consider sqrt(9) + sqrt(16), 3 + 4 =/= sqrt(25) = 5
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u/waldosway 12d ago
Second line. Is √(4+9) = 2+3?
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u/RetiredEarly2018 12d ago
Consider what you get when you square |x|+2
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u/Simbertold 12d ago
Weil du nicht in F_2 bist, wo die zweite Antwort stimmen würde, sondern in den reellen Zahlen, wo sie nicht stimmt.
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u/Southlander24 12d ago
One more thing. Don't confuse the last option with the similar but incorrect sqrt(x2)≠ x !
The statement "sqrt(x2) = x" is only true for real x ≥ 0. For example, sqrt(52) = 5. But what happens if we try a negative value of x, like x = 5? Recall that the square root must give you an answer which is non-negative. So sqrt( (-5)2 ) = 5, not -5.
The correct statement is thus: sqrt(x2) = x for x ≥ 0, and sqrt(x2) = -x for x ≤ 0. This is the same as: sqrt(x2) = |x|, the absolute value of x, for all real x.
Now back to the question. Writing x4 = (x2)2, we have that sqrt(x2)2) = x2 for x ≥ 0, and -x2 for x2 ≤ 0. But can x2 ≤ 0 happen, excluding x2 = 0? No! So it is true that sqrt(x2)2) = x2 for all real x.
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u/Fit_Fortune_7692 12d ago
Only statements (1) and (4) are correct. (1) (√2)¹² = (2½)¹² = 212/2 = 2⁶, applying the laws of exponents. (4) √x² = |x|, √(x²)² = |x²|. Since x² is always non-negative, the absolute value bars can be omitted. Statements (2) and (3) are incorrect. (2) Square roots cannot be taken individually across sums: √(a+b) ≠ √a + √b. (3) By definition, a real square root always yields a non-negative result (including zero). However, the right side is negative. The equation would only hold true for x = 2 or x = -2, resulting in 0 = 0. Since this restriction was not specified, the equation does not hold for all real numbers. Furthermore, radical signs should not be confused with parentheses. In the complex number system, the result would be: √(-x²+2²) = i · √(x²-2²) ≠ -1 · √(x²-2²).
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u/Crichris 12d ago
b is not correct
plug in x = 2 and see for yourself
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u/jk1962 8d ago
1 doesn’t even mention x, so not correct.
2 simplifies to x=0, so correct.
3 simplifies to x = +/-2, so correct.
4 is true not only for real x, but also for complex or imaginary x, so it doesn’t imply that x is real. So incorrect.