Can someone give me an explanation on how to do Completing the Squares in Algebra?
471239
Here's how to complete the square:\n\nI'll use the example of \n\n2x^2 - 4x -6 = 0\n\n1.  Divide all the terms by the coefficient of the x^2 term.\n\nThe coefficient of the x^2 term in your example is 2, so\n\n(2/2)x^2 - (4/2)x -(6/2) = 0\nx^2 - 2x -3 = 0\n\n2.  Bring the constant term to the right side of the equation.\n\nx^2 - 2x = 3\n\n(This is where you are at in your example)\n\n3.  Complete the square\n\nAny square of the form (x + a)^2 is x^2 + 2a + a^2.  Look at the x term in your example.  The coefficient is -2.  In your example, 2a is going to be -2.  So a=-1.  Right now, you have (on the right side)\n\nx^2 +2(-1)\n\nThe missing part of the square is (-1)^2, which is equal to 1.  You complete the square by adding that value to the right side.  You keep the euqation balanced by adding it to the left side:\n\nx^2 - 2x +1 = 3+1\n(x - 1)^2 = 4\n\n4.  Take the square root.  Remember that there are two possibilities:\n\nx - 1 = 2\nx - 1 = -2\n\n5.  Solve both equations:\n\nx = 3\nx = -1\n\n6.  Check both your answers.  Anybody in the world could make an error here.  The most important thing is to check both of your answers:\n\nx = 3\n\n2x^2 - 4x -6 = 0?\n2(9) -4(3) -6 = 18 -12 -6 =0\n\nx = -1\n\n2x^2 - 4x -6 = 0?\n2(1) -4(-1) -6 = 2 + 4 - 6 = 0
91424
x*SQRT(8+3x)*SQRT(50-4)*2x^2?????\n\nYour statement can be expressed as different algebraic expressions.\n\nExample: The last statement, "2x squared" can be 2x^2 or (2x)^2.\n\nTry to write it in algebraic form with, paranthesis if needed.
812516
Concentrate on the ^2, because that will have the most significant impact on the value of the expression being raised to the second power:\n\nFirst I tried just raising the 4 to the second power.  4^2 = 16, so "n" / 16 = 6, therefore n = 96.  If you can group the other numbers to reach 96, you have an answer.  But working that out for a while got me nowhere, so I started working leftwards to bring the next number and operator into the square:\n\n(6/4)^2 = 2.25\n\nThat is not an integer so it probably won't work.\n\nEven bringing the next number and operator in, which is 2 -, I still end up with a non-integer:\n2 - (6/4)^2 = -0.25\n\nBut look -- the next number and operator are "8 x " so lets work that out:\n\n8 x (2 - (6/4)^2) = 8 x (2 - 2.25) = 8 x (-0.25) = -2\n\nso that might work if we can come up with a number which works out for ___ + (- 2) = 6\n\nWhat plus -2 is =6?  It is 8 of course.  Now we just need to group 3 x 5 - 3 + 2 to equal 8:\n\n(3 x (5-3)) + 2\n(3 x (2)) + 2\n(3 x 2) + 2\n6 + 2\n8\n\nAha!  We have it.\n\nSo then, or grouping is:\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n\nTechnically some of those parenthesis are extras because the order of operations says multiplications and divisions are always done first, you don't need to explicitly show that they are (but it helps keep things clear).  So this would be a simpler form:\n3 x (5 - 3) + 2 + 8 x (2 - (6 / 4) ^2) = 6\n\nBut anyway, putting it back in the more complex form and proving it true works like this:\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n(3 x ( 2  )) + 2 + (8 x (2 - (( 1.5 ) ^2))) = 6\n(3 x   2   ) + 2 + (8 x (2 - (    2.25   ))) = 6\n(3 x   2   ) + 2 + (8 x (      -0.25        )) = 6\n(6) + 2 + ( -2) = 6\n6 + 2 - 2 = 6\n6 = 6 <<<<< TRUE!\n\n\nHere is another grouping that also works out:\n3 x (5-3) + 2 + 8 x ((2 - 6) / 4^2) = 6\n3 x    2    + 2 + 8 x     (-4 / 16) = 6\n3 x    2    + 2 + 8 x (-0.25) = 6\n3 x    2    + 2 - 2 = 6\n6 + 2 - 2 = 6\n\nand another, this one involves only integers in the proof so might be more what you are looking for:\n3 x (5-3) + 2 + (8 x (2 - 6)) / 4^2 = 6\n3 x 2 + 2 + (8 x (-4)) / 16 = 6\n3 x 2 + 2 + (-32) / 16 = 6\n3 x 2 + 2 + (-2) = 6\n3 x 2 + 2 - 2 = 6\n6 + 2 - 2 = 6\n6 = 6
291785
To find the numbers you have to solve these 2 equations simultaneously.\nThe sum of two integers is 16 translates to the equation:\nx + y = 16\nThe sum of their squares is 130 translates to the equation:\nx^2 + y^2 = 130\nSolving for x and y gives you 7 and 9.\nAlways check to see if your answer makes sense.\n7 + 9 = 16\n7^2 + 9^2 = 49 + 81 = 130
766449
(-x/4+ 5/6) * 12 < (4x/3 +2) * 12 (mulitply both sides by 12)\n\n-3x +10 < 16x + 24\n\n-3x-16x +10-10 < 16x-16x + 24-10 (move all variables and numbers to one side)\n\n-19x< 14 (divide both sides by 19)\n\nx < -14/19\n\n* means times\n\n(-n/5 + 6) * 5 > n *5 (multiply both sides by 5)\n\n-n + 30 > 5n\n\n-n + n +30 > 5n +n (move all varialbes to one side)\n\n30 > 6n (divide both sides by 6)\n\n5>n
789616
Since the diagonals of a square bisect each other, 2x - 1 = 3y + 5.\n\nSince the diagonals are congruent, 4x - y + 1 = (2x - 1) + (3y + 5)\n\nGet x and y on the same side of = in each:\n\n2x - 3y = 6\n\n2x - 4y = 5\n\nMultiply the bottom row by -1 and add it to the top to eliminate x:\n\n2x - 3y = 6\n-2x+4y = -5\n\ny = 1\n\nPlug this in to an equation to find x:\n\n2x - 3 = 6\n\n2x = 9\n\nx = 4.5
685700
First convert into a matrix form which would be\n\n1    1    1\n1    2    3\n1    3    7\n\np\nq\nr\n\n= \n\n1\n4\n13\n\nTake the square matrix (3x3) on the other side so u ll need its inverse (i have directly calculated the inverse which is)\n\n5/2       -2      1/2                        \n-2         3      -1                           \n1/2       -1     1/2   \n\nMULTIPLY BY    \n\n1\n4\n13\n\n& the result is\n\n\n 1        p\n-3        q\n 3     =  r\n\nthis is the answer of the equation
243121
first square two sides\nyou will get 3x+1 =9+6√(x-4)+x-4\n= (x-2)=6√(x-4)  square it \nyou will get x^2 -4x +4=9(x-4)\n=x^2 -13x + 40=0\nfactor  (x-5)(x-8)\nso x=5 or x=8
780960
I also have difficulty typing this thypes of math problems.  First, simplify the ax^2+bx+c parts in the first half to (x+c)(x+c) looking things first.  So...\n\n[(x-3)(x-1)]} is the first part.  With the bottom of the second half, divide by 3, so that you get 3(x+1).  Now line everything up like so...\n\n(x+1)(x+3)(x-3)\n_____________\n3(x-3)(x-1)(x+1)\n\nNow do the cancellations and you get the right answer...\n(x+3)\n____\n3\n\nSorry for not working to convention, but it was easier to type it that way.
128596
im great with algebra, ive been doing it for the past 4 years now non stop, email me and ill be more than happy to show you how to do it step by step and answer you questions\nzinger_38@yahoo.com
832845
you need to add +1\nso that the equation can be represented as a perfect square\nx^2 - 2x + 1 = (x - 1) (x - 1) = (x - 1)^2\n\nthis is called solving by perfecting the square\nto do this, make sure that the coefficient of the first term is 1.\nThen, divide the coefficient of the middle term by 2,\nthen square the whole thing.\n\nIn this case, the coefficient of the middle term is -2\n-2 / 2 = -1\n(-1)^2 = +1
837436
Your Algebra teacher is trying to kick your butt, right?  I have the answer:\n\nReorder the terms:\n3+64x+16x^2=0\n\nStart with the 16 - divide all numbers by 16\n\n3/16+64x/16+16x^2/16=0/16\n\nResults are:\n\n.1875+4x+ x^2=0\n\nNot done yet...Next, take out the .1875.\n\n.1875+(-.1875)+4x+x^2=0+.1875\n\nResults from this calculation:\n\n4x+x^2=-.1875\n\nThen, add a 4 into the equation\n\n4+4x+x^2=-.1875+4\n\nAdd like terms:\n\n4+4x+x^2=3.8125\n\nThen, you need to factor the problem out.\n\n(x+2)(x+2)=3.8125\n\nTake the square root of 3.8125\n\n=1.952562419\n\nAfter that, take one of the factors you created above and use the square root to arrive at the answer...one positive, one negative\n\n2 + x = 1.952562419\n\nx = 1.952562419 - 2\n\nx = -.0474437581\n\nand\n\n2 + x = -1.952562419\n\nx = -1.952562419 - 2\n\nx = -3.952562419\n\nx = (-.047437581, -3.952562419)\n\nYou might have to plot that on a graph.  You have just calculated a point of off the x axes...the horizontal line.\n\nHope that helps. Maybe someone can give you fewer steps.
385232
If you're talking about algebra I, you are probably dealing with quadratic expressions like the following:\n\nx^2 + 7x + 6\n\nThese are a bit like a puzzle.  If the number in front of the x^2 term is 1 (as in the above example) then the puzzle is like this:\n\nLook at the 7 and the 6.  What two numbers multiply to 6 and ad to 7?\n\nConsider all the pairs of numbers that multiply to 6\n\n1,6 or 2,3.  Which pair adds to 7? The 1 and 6.\n\nSo the above equation factors to (x + 6)(x + 1)\n\nAnother example is x^2 - 6x - 16\n\nThe ques ion becomes what two numbers multiply to -16 and add to -6.\n\nThe pairs of factors of -16 are \n+ or - 1 and 16\n+ or - 2 and 8\n+ or - 4 and 4\n\nWhich pair of these factors add to -6?  With a little checking you can see the pair is -8 and +2\n\nSo this equation factors to\n(x - 8)(x + 2)\n\n\nThis even works for equations like the following:\nx^2 - 16\n\nWhat you have to remember is that this can be re-written as\n\nx^2 + 0x -16\n\nSo, what multiplies to -16 and add to zero.  Use the last example and you should see this one for yourself.\n\nGood 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
84611
Check this website:\nhttp://mathworld.wolfram.com/Addition-MultiplicationMagicSquare.html
268070
Check out this site. It gives you a few examples. It was helpful for my cousin when she was leaning how to do them.\nhttp://www.purplemath.com/modules/solvelin.htm
981742
x² + 2x + y² - 6y - 6 = 0 --- Add 6 to both sides...\nx² + 2x + y² - 6y = 6 --- Add some parentheses...\n(x² + 2x) + (y² - 6y) = 6 --- Complete the Squares*...\n(x² + 2x + 1) + (y² - 6y + 9) = 6 + 1 + 9 --- Simplify the two expressions in the parentheses, and add the numbers to the right of the "=" sign...\n(x + 1)² + (y - 3)² = 16\n\nFormula: (x - h)² + (y - k)² = r² --- Where (h, k) is the center point of the circle, and r is the radius of the circle.\n\nUsing this formula, (-1, 3) would be the center, and 4 (= √16) is the radius.\n\n\n*Completing the square is an algebra technique, also used in many types of calculus. The essential objective is to reduce a quadratic polynomial in a variable in an equation or expression to a squared polynomial of linear order. This makes the equation easier to solve in many contexts. More information can be found here: http://en.wikipedia.org/wiki/Complete_the_square