If 2 is the square root of 4, what is the square root of minus 1 and why would you care?
461173
Maybe in mathematics sqrt of -1 = i, but in electrical engineering it is j.\nIn electrical engineering and related fields, the imaginary unit is often written as j to avoid confusion with a changing current, traditionally denoted by i.\n\nThis represents the imaginary component of electronics which cannot be ignored. Phasor diagrams may help to explain this phenomenon. A phasor is a constant complex number representing the complex amplitude (magnitude and phase) of a sinusoidal function of time. It is usually expressed in exponential form. Phasors are used in engineering to simplify computations involving sinusoids, where they can often reduce a differential equation problem to an algebraic one.
747566
-1.4172^2=\n\n-1.4172 times -1.4172.\n\nA negative times a negative is always positive, so you just multiply them together to get a positive value.  When you do so, you get a number a little larger than 2.  The square root of two (the number that, when multiplied by itself, equals 2) is about 1.414.... Your number is slightly larger, so the answer will be slightly larger than 2.\n\nHope this helps.
981873
-7 and -sqrt49\n\nYou can not take a square root of a negative number. Well, you can, but it's called an imaginary number because it can't actually exist.
1116021
Cannot be, because any whole number ending on a 1 squared will end on a 1.\n\nDid you mean the square root of 12345678987654321?\n\nThis results from summation:\n\n111111111^2 multiplied manually is\n\n11111111100000000\n01111111110000000\n00111111111000000\n00011111111100000\n00001111111110000\n00000111111111000\n00000011111111100\n00000001111111110\n00000000111111111\n\n12345678987654321
941657
Nah, sorry, but something in your thinking is incorrect.  You seem to be saying that if you add the square root of 3 four times, you want to get 9.  That's not right.  If you take sqrt(3)^4, you WILL get 9.   Taking the fourth root of a number is not like dividing a pie into four pieces.
1031543
There is no exact square root but it is\n\n7414.85009963114
24322
2.4 (n - 1) = 12\n->(n-1)=12/2.4\n->(n-1)=120/24\n->(n-1)=5\n->n=6\n\nKeep Asking
593847
The answer is zero isn't it?  The square root of (7 squared) is 7.  If you minus 14 you get -7. Add another 7, brings you up to zero.
337853
The answer, before simplification, is sqrt(-8). There is no need to panic, as some of the other answerers did. This simplifies easily to\n\n2 * i * sqrt(2)
862673
since there is no real answer to the square root of any negative number we write i\ni=the square root of -1\n\nif you did the square root of -25 you would get 5i which equals to 5 times the square root of -1
1058646
x= -b +/- square root of  b2 - 4ac all over 2a\nso the answer is x=-.7 and x=4.3
435410
2i or -2i, where i = sqrt(-1)\n\nIf you haven't learned imaginary numbers, then your answer would be that the sqrt(-4) is not a real number.
678825
The square root of -1\nis an imaginary 1\nit is depicted generally with a character 'i'\nif u want root of -2\nthen itd be i x root(2)\nor 1.414i
622009
no. square root 32 = square root 4*4*2 = 4 square root 2\n\nbreak down the 32 into parts that will come out of the root. Then if there are two numbers inside, they become 1 number outside. If there is a loner number inside, then it stays there.
737743
The positive square root is 11.  The cube root is about 4.946 (it's not an integer).
660883
The square root of 4/25 is 2/5, and that is a real number.  The square root of any positive number will be real.\n\nHowever, sqrt(-121) does not have a real solution; it will be complex because it is the square root of a negative number.  In this case, it would be 11i.\n\nThe solutions to x^2 = -121 are x = 11i and x = -11i.
498746
Square root of four has two solutions: 2 and -2.\n2 is a natural number. All natural numbers are integers. All integers are rational.\n-2 is NOT a natural number (no natural number is negative). But it is an integer, so it's a rational number.\nAs rational numbers are opposed to irrational numbers, none of the two solutions can be irrational.