sqr root of:  a^2+b^2 =a+b?
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.
541116
3=2 did get a correct answer...maybe you need to review your rules of square roots.
672272
I have taken the liberty of calling it the (a+b)^2=a^2+b^2+2ab ; (a+b+c)^2=a^2+b^2+c^2+2ab+2ac+... ; (a+b+c+d)^2=a^2+b^2+c^2+d^2+2a...\n+2cd etc... theorem. A bit long winded but exact.
1029855
D\n\nSeriously, what class are you in? Algebra? Geometry? You chould know this!! It's the pythagorean theorm- [a2 +b2 =c2] (2's = squared).  You have a and b, the two adjacent sides of the triangle.  Therefore, it is the square root of 50, or approx. 7.07.  Thats SUPER important, make sure you remember it!
972642
(A^2+2ab+b^2)/(a+b) + (a^2-b^2)/(a-b) + (a-b)/(a+b)/(a^2-b^2)/a+b)\n(a+b)^2/(a+b)+(a+b)(a-b)/(a-b)+ (a-b)/(a+b)*(a+b)/(a^2-b^2)\na+b+a+b+(a-b)/((a+b)(a-b))\n2a+2b+1/(a-b)\n\nif I got all the ( ) in the right places.
411391
y=(4b/a^2)x^2\n 1. sqaure root both sides to ge sqrt(y)=sqrt((4b/a^2)x^2)\n 2. the 2 in the numerator 2^2=4, the a in the denominator and the x comes out from under the root leaving the b under root so we have sqrt(y)=(2/a)*x*sqrt(b)\n 3. multiply both sides by (a/(2*aqrt(b)) to solve for x\n 4. x= + or -(a/(2*sqrt(b))*sqrt(y) two cases !!\n 5. plot the + and - solution seperately to obtain a parabola rotated 90 degrees clockwise
1001448
[sqrt(2)]^2 = c^2 - b^2\nLet c = 3/2 and b = 1/2, then\n[sqrt(2)]^2 = (3/2)^2 - (1/2)^2\n2 = 9/4 - 1/4\n2 = 8/4\n2 = 2
166521
Yes it is; and let us see how:\n\n(a+b)square = (a+b) x (a+b)\n= a x a + a x b + b x a + b x b\n= a x a + a x b + a x b + b x b\n= a square + 2ab + b square\n\nsimple, isnt it?
1049820
The quadratic formula:\nx= -b(+/-) the square root of (b squared - 4ac), all of this over 2a
139883
(a+b)(a-b) \n= a^2 - ab + ba - b^2\n= a^2 - b^2 -ab +ab\n= a^2 - b^2
980481
Use the Pythagorean Theorem of:\na² + b² = c²\n\na and b are legs, c is the hypotenuse.\n\nPutting in the numbers you know:\n7² + b² = 13²\n\nNow solve for b:\nb² = 13² - 7²\nb² = 169 - 49\nb² = 120\n\nTake the square root:\nb = sqrt(120)\nb ≈ 10.954
557831
Yes. You are right, but if and only if the equation MUST be valid for all a & b. The reason is:\n(x+a+b)^2 - x^2 = (x+a+b - x)*(x+a+b + x) = (a+b)*(2x+a+b)\nand\n(x+a)^2 - (x+b)^2 =[x+a -(x+b)]*[x+a + (x+b)]= (a-b)*(2x+a+b)\ncomparing the above, and noting that a+b could not be equal to a-b for all a & b, then 2x+a+b=0 so x=-(a+b)/2\nSimilarly the second equation could be solved by giving square root of both sides:\nx+a = +/-(x+b)\nit is obvious that "+" sign could not be selected since it required that a=b which is against our assumption. So:\nx+a=-(x+b) ==> x=-(a+b)/2
772527
x^2 + 3x  = -1\nx^2 + 3x +(3/2)^2 = -1 + (3/2)^2\n(x +(3/2))^2 = (5/4)\nx + 3/2 = sqrt(5)/2 OR x + 3/2 = - sqrt(5)/2\n\n(sqrt means the positive square root) \n\nx = -(3/2) + (1/2)sqrt(5)\nx = -(3/2) - (1/2)sqrt(5)\n\n.......................................................................\n\n2x^2 +4x + 1 = 0\nx^2 + 2x = -(1/2)\nx^2 + 2x + 1 = (1/2)\n(x+1)^2 = (1/2)\n\n(x + 1) = sqrt(1/2) OR (x + 1) = -sqrt(1/2)\n\nx = -1 + sqrt(1/2) OR\nx = -1 - sqrt(1/2)\n....................................\nLW = 783\n\n\n\n(W+2)W = 783\nW^2 + 2W = 783\nW^2 + 2W  + 1 = 784\nW + 1 = sqrt(784)\n\nW = srqt(784) -1\nW + 2 = sqrt(784) + 1\n\n2(W + L) = 2(2) sqrt(784)\n                = 4* 28\n                 =   112
752173
Sqrt(3x/2) = Sqrt(3x/2) * SQRT(2/2) = SQRT(6x/4) = 1/2 * SQRT(6x)\n\n6*SQRT(7) - SQR(28) = 6*SQRT(7) - SQR(4*7) = 6*SQRT(7) - 2*SQR(7) = 4 * SQRT(7)
600301
x^3+x^2+2 = x^5+2\n\n(x^2+2)^2 = x^4+4\n\n(x^3+x^2+2)/[(x^2+2)^2] = (x^5+2)/(x^4+4) = (x+2)/4 = (x+1)/2
1066820
If 'a' is a root of 2x^2=3x-4, then 2 a^2=3a - 4.\ni.e. a^2=(3a - 4)/2\n\nNow consider 4 a^3\n4 a^3\n= 4a(a^2)\n= 4a(3a - 4)/2\n= 2a(3a-4)\n= 6a^2 - 8a\n= 6(3a-4)/2 - 8a\n= a - 12\nQED
267409
(-b + or - the square root of (b squared - 4AC)) divided by 2A