If s is a variable and you square it ,and get the sqrt of S^2 would you get  the same value of S back? Why?
612102
you DONT get the same value back !\n\nyou get TWO different values back whenever you obtain the square root of ( s^2 ).  for example the sq rt of 25 is 5 or (-5) .\n\nthus, solution to a square root always involves TWO values ( +/- ).\n\nin other words, "s"  can be positive or negative only but the solution is not unique ( +s or -s ) and unspecifiable.\n\nthanks gore .... :)
999326
If you start with a number bigger than one, then the square root will always be smaller than the number, but bigger than 1.  So every time you take the square root, the number gets smaller, but never gets smaller than one.  Eventually it's so close to one that the calculator just displays it as 1.
456165
yes. for example, 8 times 8 is 64...so 8 squared is 64 and the square root of 64 is 8
353701
sqrt(a^2 + b^2) = a + b\nSquare each side.\n[sqrt(a^2 + b^2)]^2 = (a+b)^2\na^2 + b^2 = a^2 + 2ab + b^2\nSubtract a^2 and b^2 from each side\n0 = 2ab  \nDivide both sides by 2\n0 = ab\nEither a = 0 or b = 0 or both equal 0.\nAnswer: a=0 and b > 0, or a > 0 and b=0, or a=0 and b=0\na or b can never be negative(even it one of them is zero).\n\nIn general, the equation \nsqrt(a^2 + b^2) = a + b \nis false.  However, it is true for certain values of a and b.
881442
It is false no solution\n\nLook at \n Example 5:  on this site\n http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut14_solve.htm\n \nWhere did our variable, y, go???  It disappeared on us.  Also note how we ended up with a FALSE statement, 0 is not equal to 11.  This does not mean that y = 11 or y = 0.  \nWhenever your variable drops out AND you end up with a false statement, then after all of your hard work, there is NO SOLUTION. \n\nSo, the answer is no solution.
698714
Assuming you mean S=(c-b)/(b-a) rather than c-1-a...\n\nMultiply both sides by (b-a) to get S(b-a)=c-b\nMultiply the S through to get Sb-Sa=c-b\nAdd Sa+b to both sides to get Sb+b=c+Sa\nFactor out b to get b(S+1)=c+Sa\nDivide both sides by (S+1) to get b=(c+Sa)/(S+1)\n\nKeep in mind this only holds true for b not equal to a because then S would have no value.\n\nIs that what you're looking for?
798185
Yes, they do.  Domain just means the values the relation works on, range means the values it produces.\n\nThe big deal about functions is that they give you a definite, i.e. unambiguous result.  If you ask what do I have to square to get 324, the answer could be 18 or -18.  But put it into your calculator and press the square root key, and it operates the function (which is defined as the positive value of the two choices) and it gives you 18.\n\nLikewise, there's an infinite number of answers to the question "What angle (in degrees) has a cosine equal to -0.5?"  A few of the answers are 120, 240, 480, 600.  But again, put -0.5 into your calculator, set in degrees, and apply the "cos^-1" key, and your calculator applies a function and comes up with just one answer, 120.
700071
Yes. If you multiply it out it is the same. The square kills off that minus sign because any real number squared is positive.\n\n\nz
588403
y not.. x^4+y^4+2X^2y^2 is the square of x^2+y^2..\nif u had meant square root... then sqrt(x^+y^2) is sqrt((x+y)^2-2xy))
842246
y = (x + 2)^2\n\nsqrt(y) = x + 2\n\nx = sqrt(y) - 2\n\nSo, finv(x) = sqrt(x) - 2.\n\nIt is limited to the domain x>=0.
802534
I'll help set it up\nbut, honesty, it isn't that hard and the only way you'll learn it is to do some on your own so that at test time, you can do it yourself...\nI'll start your homework....you finish....\n\nthe essential idea is to complete the square for each variable\nfirst reorder the terms \nx^2 - 2x +____  - 9y^2 - 18y + ___ = 9\nfactor out where possible\nx^2 - 2x + ___ - 9 (y^2  + 2y + ___) = 9\ncomplete the sq for both terms\nin case you were wondering why I put a "space" for the constant term in the working\nx^2 - 2x + 1  - 9(y^2 +2y + 1) = 9 + 1 + (-9)\nfactor\n(x - 1)^2 - 9(y+1)^2 = 1\n\nOK?????\nyou take it from here...it's almost in standard form
979217
y = sqrt(1+t^2) = (1+t^2)^(1/2)\n\nby the power and chain rule\n\ndy/dt = (1/2)(1+t^2)^(-1/2) x d/dt (1+ t^2)\n\n= 2t(1/2) /sqrt(1+t^2)\n\n= t/sqrt(1+t^2)\n\nTherefore\n\ndy = (t / sqrt(1+t^2)) dt
1152199
It doesn't have a square root. A square root has to be an exact number (no decimals). Because it means that it is whatever that number is squared. ex: 100=10^2(10 squared). If it's a decimal number that goes on forever, it's impossible to square it because it's kinda infinite.
803771
sq.rt of x*y = Sq.rt X * sq.rt. Y\n\nIt can be proven that:\n √(x*y) = √x * √y\n\nConsider the first part → √(x*y).\nTwo number values, one is x and the other is y, they are multiplied together.  You then take the square root of this result.\n\nConsider the second part → √x * √y.\nYou take the square root of x and y and then you multiply the results together.\nIt can be shown that the results of the first part is the same as the result of the second part.
737774
You figure the square root of each number separately...you've probably learned that in class alredy (how to do that)...then you take reach result, and multiply them together for your final answer.\n\nUse your textbook for reference...good luck!
523409
say you had x^2 + 4x + 3 = 0\nto complete the square you want the right side to be "squarable." i.e. in the form (x+a)^2 = b\nin this case the form is\n(x+2)^2 = x^2 + 4x + 4\n\nso you add one to both sides and yet\nx^2 + 4x + 4 = 1\n\nthis equals\n(x+2)^2 = 1\nx+2 = sqrt(1) = +/-1\nx therefore = -3, or -1
953913
It's a difference of squares...\nIf you have (a² - b²) that is equal to (a + b)(a - b)\n\nSo in your example:\n16 - 9x²\n\n16 = 4² = a²\na = 4\n\n9x² = (3x)² = b²\nb = 3x\n\n(a + b)(a - b)\n= (4 + 3x)(4 - 3x)