What is the Hindu Method of Completing the Square?
913819
The "Hindu Solution" to the quadratic equation is the same method that we now call "completing the square."  This is a way to convert terms in the equation to the form of the square of a different polynomial, enabling it to be solved algebraically.  When applied to a quadratic equation in the general form, this method yields the quadratic formula.
874062
This progression is created by squaring the previous number and adding 1.\n0 squared = 0, add 1 and you get 1.\n1 squared = 1, add 1 and you set 2\n2 squared = 4, add 1 and you get 5,\nand so on.
1087500
Just to complete to previous answer, in case it isn't clear:\n\nTo make it a perfect square you add (-b/2a)^2 to both sides.\nAs noted: \n(-7/2)^2 = 49/4\n\nStarting with:\nx^2 + 7x = -12\n\nAdd 49/4 to both sides:\nx^2 + 7x + 49/4 = 49/4 - 12 = 1/4\n\nFactor the perfect square:\n(x + 7/2)^2 = 1/4\n\nNow take the square root of both sides remembering that there are two roots (+ and -):\nx + 7/2 = +/- 1/2\n\nSo there are two equations\nx + 7/2 = 1/2  or x + 7/2 = -1/2\n\nSo x = -3  or x = -4
880996
No.\nTo simplify SQR(108), break it down into\nSQR(36)*SQR(3)\nSince 36 is a perfect square, SQR(36) = 6\nSo the answer is 6*SQR(3).
1040461
find the area of the half circle first\n\nthen find the area of the round part of the yin yang that sticks out. that circle find the area of the half circle . and then subtract that area from the big half circle area
741830
the ratio is 1:2 or 1/2\n\noh, by the way it goes like this:\n\nassuming the length of each side of square A is a, then the area is a^2.\n\nsince, the vertices of the smaller square (A) is the midpoint of the sides of the big square(B), using the diagonal of square A, we can find the sides of the bigger square.\n\nThe diagonal is a*squareroot of 2, in which when we square to get the area ===>  gives us 2a^2.\nThus, the ratio is between Square A' area (a^2) and B's area (2a^2) is 1:2 or 1/2
751528
Yes.  For example, take the two from the top left corner, and the two from the bottom right corner, and make a new square touching the top right corner.\n\n\nAAAA BBBBB\nC.........D..........E\nC.........D..........E\nC.........D..........E\nFFFFFF GGGG\nH.........I............J\nH.........I............J\nH.........I............J\nKKKKK LLLLL\n\nbecomes\n . . . . . . . . . . . . . AAAA\n . . . . . . . . . . . . .C . . . . J\n . . . . . . . . . . . . .C . . . . J\n . . . . . . . . . . . . .C . . . . J\n. . . . . . .BBBBB LLLLL\n. .........D..........E\n. .........D..........E\n. .........D..........E\nFFFFFF GGGG\nH.........I\nH.........I\nH.........I\nKKKKK
258864
there is a method of getting the square root of any number!!\n\nfirst, group the digits in two's, starting from the decimal pt.\n\n1'23'45'43'21.00'00\n\nThen, do the ff:\n1.Get the sq.root or the nearest sq.root of the first group of digits (sqrt3=1, sqrt5=2), that is the first digit of your square root.\n2. Square that number and subtract it from the first group of digits. Bring down the first digit of the next pair of digits.\n3. Multiply the digits of your sq.root by 2 and use this as the divisor of the numbers you brought down. this will be the next digit of your sq.root. Put it next to the previous digit.\n4. then, multiply, and subtract, just like in division. Bring down the next digit of the pair.\n5. Square the so far last digit of the sq.root, and subtract this from the brought-down number.\n6. Repeat the steps until the right decimal pt.\n\nThus,\n\nsqrt1'23'45'43'21\n\nsqrt1 = 1 (1st digit)\n1x1 = 1\n1 - 1 = 0\nBring down 2\n\n1x2 = 2\n2/2 = 1 (2nd digit)\n1x2 = 2\n2-2 = 0\nbring down 3\n1² = 1\n3-1 = 2\nBring down 4\n\n11x2 = 22\n24/22 = 1 (3rd digit)\n1x22 = 22\n24-22 = 2\nbring down 5\n1²=1\n25-1=24\nBring down 4\n\n111x2=222\n244/222=1 (4th digit)\n1x222=222\n244-222=22\nBring down 3\n1²=1\n223-1=222\nBring down 2\n\n1111x2=2222\n2222/2222=1 (5th digit)\n1x2222=2222\n2222-2222=0\nBring down 1\n1²=1\n1-1=0 STOP..\n\nThe answer is 11111.^_^
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!
602085
You use Hero's formula. Add together the sides and divide by two (also called the semiperimeter), and let that value be S. Arbitrarily pick sides to be A, B, and C, then do the following:\n\nArea= Square root (S*(S-A)*(S-B)*(S-C))\nAnd you'll have the area.\n\nIncidentally, Brahmagupta's formula works the same way for quadrilaterals that can be inscribed in a circle, only you use A=square root (S-A)*(S-B)*(S-C)*(S-D). S remains the semiperimeter.
461438
The order of mathmatical operations can be abbreviated as BEDMAS\n\nBrackets\nExponents\nDivision\nMultiplication\nAddition\nSubtraction\n\nYour question can be written SQRT (144 x 12 - 4)\n\nDo the brackets first, again following BEDMAS\n=1724\nThen the SQRT which is essentially an exponent.\n=41.52....\n\nIt's so nice to see so many eager to answer the question wrong.
1137789
Here are some of the things you can do to a whole number!\nNote: These can be done using a calculator\n\n1)\nAlso get the number next to it. (e.g. if you chose 5, get 6)\nmultiply them.\nmultiply the result by 4.\nadd 1.\nget the square root of it.\nSubtract 1.\nDivide by 2.\nThe result is the original number!\n\n2)\nSquare your number.\nAdd twice the original number\nAdd 1\nGet the square root\nSubtract 1.\nThe result is the original number!\n\n3)\nSquare your number.\nAdd 6 timesthe original number\nAdd 9\nGet the square root\nSubtract 3.\nThe result is the original number!\n\n4)\nSquare your number.\nMultiply by 4.\nAdd twelve times the original number\nAdd 9\nGet the square root\nSubtract 3.\nDivide by 2.\nThe result is the original number!\n\n5)\nNote: Here, you must choose a single digit whole number.\nmultiply it by 37.\nAdd the digits of the result.\nContinue adding the digits of the result until you get a single-digit result.\n(e.g. 5 x 37 = 185, \n1 + 8 + 5 = 14, \n1 + 4 = 5)\nThe result is the original number!\n\n\n6)\nAlso get the next two consecutive whole numbers.\nmultiply the first and the third.\nsubtract the square of the middle term.\nThe answer is always 1!\n\n7)\n(note: you must choose here a 3-digit whole number)\nmultiply it by 7\nmultiply it by 11\nmultiply it by 13\nThe result is the number repeated!\n\n8)\nnote: you must choose a single-digit whole number\nAdd 3\nMultiply by 2\nsubtract 1\nMultiply by 5\nAdd 580\nRemove the tens digit of the result.\nThe result is always 65.\n\nFrom here, you will no longer choose a number\n9)\n9 x 9 + 7 = 88\n9 x 98 + 6 = 888\n9 x 987 + 5 = 8888\n9 x 9876 + 4 = 88888\n9 x 98765 + 3 = 888888\n9 x 987654 + 2 = 8888888\n9 x 9876543 + 1 = 88888888\n9 x 98765432 + 0 = 888888888\n\n10)\n1/7 = 0.142857 142857 142857 ...\n2/7 = 0.285714 285714 285714 ...\n3/7 = 0.428571 428571 428571 ...\n4/7 = 0.571428 571428 571428 ...\n5/7 = 0.714285 714285 714285 ...\n6/7 = 0.857142 857142 857142 ...\nNotice that each is a repeating pattern of 142857\n\n^_^
844185
square-ish in shape\ncome in different sizes\nmany varieties
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
1130245
I used to be a square but I'm all _right_ now :-)\n\nSeriously, you should do your own homework.
708943
You have to complete the square.\n\ny = a(x^2 + b/a x) + c\ny = a(x^2 +b/a x + b^2/4a^2) + c - b^2/4a\ny = a(x + b/2a)^2 + (4ac - b^2)/4a\n\n[4ay - (4ac - b^2)]/4a = a(x + b/2a)^2\n[4ay - 4ac + b^2]/(4a^2) = (x + b/2a)^2\n\n+/- sqrt(4ay - 4ac + b^2)/2a = x + b/2a\n\nx = (-b +/- sqrt(4ay - 4ac + b^2))/2a
572512
hey,\nanswer is go for faster mathematics-vedic maths\ncheck out following link-it would help u lot\nlink:-  http://www.sanalnair.org/articles/vedmath/example1.htm\ntry all examples in the link\n\nhey purwa,u gonna give me 10 points 4 this(vote me best)