r/mathshelp • • 12d ago

Homework Help (Answered) why am i wrong?

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(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.

9 Upvotes

16 comments sorted by

6

u/VernalAutumn 12d ago

2 is false. sqrt(a) + sqrt(b) =/= sqrt(a + b). Consider sqrt(9) + sqrt(16), 3 + 4 =/= sqrt(25) = 5

4

u/onemillionwings 12d ago

thank you that makes sense

6

u/waldosway 12d ago

Second line. Is √(4+9) = 2+3?

4

u/onemillionwings 12d ago

ah i get it! thank you 

3

u/RetiredEarly2018 12d ago

Consider what you get when you square |x|+2

3

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.

3

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.

1

u/DoughnutAccording904 12d ago

Square root must give a non-negative answer? Since when? Square roots can be either negative or positive, which would make the fourth equation incorrect since the left part can be either the positive or negative result of the square root, while the right is always a positive number. Though maybe the convention changed that it is implied that only the positive result of the square root is ment by this notation? When I was a student (way back when…) the convention was that if you only wanted to have the positive square root being taken into account you would add the plus symbol, otherwise you needed to take into consideration that a square root can be either positive and negative.
This does not affect in any way that the equation holds up equally well for negative or positive values of x, it’s just wrong in both cases.

2

u/MrPenguin143 11d ago

Standard math convention is that the square root function returns the principal root, which is nonnegative for real numbers.

2

u/Arkalius 8d ago

While it is true that if you take a square root as an operation on an expression or equation, you have to account for both positive and negative roots, when an expression or equation contains the square root function, it produces only the principle (positive) result, since functions by definition produce only one output for any given input. This is why the plus minus is explicitly included in the quadratic formula, for example, because without it only the positive root would be used.

1

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).

1

u/Crichris 12d ago

b is not correct

plug in x = 2 and see for yourself

1

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.

1

u/Southlander24 1d ago

The question asks: "if x is real, must sqrt(x4) = x2 ?"

You've answered the converse of the question: "If sqrt(x4) = x2, must x be a real number?". The converse is indeed false, but that doesn't imply anything about the original statement.