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There was a problem on a test and the last step got you to
√(3+x) = (-4) I decided that the answer was 13 because when solving if you square both sides (like the textbook says) you get (3 + x) = 16. My teacher says this is wrong because (-4) isn't a real square root, and that the correct answer was "no real numbers". Also, in the book it says that each positive number has two square roots, a positive and a negative. He says if the answer in the problem had been 4 or (-4) I would have been right. I argued that if you can show the positive root without the negative one, why not vice versa?! Apparently it goes without saying. I need help because if I can prove myself right I get the points for the test (majorly improving my GPA). Please nothing that proves me wrong! Please and thank you!

2007-04-16 18:00:03 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

Parentheses are to keep everything neat. It's a teacher thing

2007-04-16 18:27:01 · update #1

5 answers

Okay, if you got to some point where you had "for a fact" that

√(3 + x) = - 4

then, yes, x = 13 should have been an legitmate answer, since when finding the square root of a number, as in √x, the answer has both signs, ±. If, instead, you got to some point where

(3 + x)² = - 4

then your teacher would have been right, this time only a complex number for x would have satisfied this equation, namely x = -3 ± 2i.

If you were doing some graph work, like finding tangent lines to a circle in the cartesian plane, the equation you've given would expect to crop up, and it would in fact be important to be aware of that √(3 + x) = ± 4, as for example it could refer to two different intersection points somewhere, not just one.

You should press your case, since I think your teacher has got this one mixed up.

Addendum: skg's answer was strange. Let's see, if -4 = (2i)², then if √(3 + x) = (2i)², we have (3 + x) = (2i)^4, which is still 16! skg's analysis makes sense if we had the following: (3 + x)² = (2i)², in which case we have (3 + x) = ± 2i. I suspect that this sort of thing is very easy to get mixed up, so continue to press your case with your teacher.

Addendum 2: Very often, unfortunately, when teaching about functions in class, the function y = √(3 + x) is frequently not shown as the true inverse function of y² - 3 = x, which is a parabola yielding a real value x for all values of y. Instead, the function y = √(3 + x) is depicted only for positive real y. Again, it depends on the problem, but for some graphical problems, you do need to be aware of the ± sign, a frequent source of mistakes when overlooked. For the teacher to be promoting the view that "only the positive sign should be considered" instead of ± is just teaching a bad habit in mathematical thinking.

2007-04-16 18:19:38 · answer #1 · answered by Scythian1950 7 · 0 1

Certainly the sqrt(16) = +4 or -4
x= 13 satisfies the equation √(3+x) = (-4).
I don't know why parentheses are around the -4. Does this have some significance to your teacher?

Clearly, -4 is one of the square roots of 16.

If those parentheses place some kind of restriction on the equation, then perhaps your teacher is correct, but I certainly do not know of any such notation. Ask him what the parentheses mean. If they were not there, would your answer be correct? Ask him to show you where such notation is explained. If he can't, I believe your answer should be marked as correct.

2007-04-16 18:23:20 · answer #2 · answered by ironduke8159 7 · 0 1

Well, if the test said to solve for x where (3 + x) = (-4)^2

then x would correctly be 13.

Does your book say anywhere that the radical symbol indicates the "principal" square root? That would be roots not less than zero.

2007-04-16 18:25:30 · answer #3 · answered by Mark 6 · 0 1

I am sorry, but your teacher is right.
Let us see how.
The equation is
sqrt(3+x) = -4 = 4.i^2 (i = sqrt(-1))
sqrt(3+x) = sqrt(2^2 . i^2)
3+x = 2i
Therefore x is not a real number.

2007-04-16 18:27:05 · answer #4 · answered by skg 2 · 1 0

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2016-12-26 10:54:58 · answer #5 · answered by graney 3 · 0 0

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