how do you solve 25x^2-49y^2    thanks?
141598
If we break down 25x^2, we get 5x.\n\nIf we break down 49y^2, we get 7y.\n\nThe resulting equation is:\n\n(5x + 7y)(5x - 7y)\n\nSolve it from there.
810768
Easiest way is to write them as two equations equal to zero.  Then equate them...\n\nStarting with your first example:\nx² + y² - 4 = 0\nx² + (y-6)² - 25 = 0\n\nNow equate these:\nx² + y² - 4 = x² + (y-6)² - 25\n\nExpand out the (y-6)²\nx² + y² - 4 = x² + (y²-12y+36) - 25\n\nNow cancel like terms (like x², y², etc.) which are on both sides:\n-4 = -12y + 36 - 25\n\nNow simplify:\n12y = 36 - 25 + 4\n12y = 15\ny = 15/12\ny = 5/4\n\nNow use the value you found (y=5/4) and solve for the other variable using one of the original equations:\nx² + (5/4)² = 4\nx² + (25/16) = 4\nx² = 64/16 - 25/16\nx² = 39/16\nHere you need to take the sqrt and you'll have two answers:\nx = sqrt(39/16)\nx = +/- sqrt(39)/4\n\nSo your two points of intersection for the first pair of equations are:\n(sqrt(39)/4, 5/4) and (-sqrt(39)/4, 5/4)\n\nThe same method works for the other problems...
57070
Starting with\nx^2 - 5x - 6.25 = 0\n\nMOVE the lone term to the right side of the equation.\n\nSo\n x^2 - 5x = 6.25\n\nTo comlete the square on the left:\nDIVIDE the coefficient of the x term (5) by two (2.5).  Then SQUARE it.\n\nADD \n 2.5^2 (which happens to be 6.25) to both sides\n\nNow\n x^2 - 5x + 6.25 = 6.25 + 6.25\n\nREWRITE the left which is now a perfect square\n (x-2.5)^2 = 12.5 (or 25/2)\n\nTAKE THE SQUARE ROOT OF BOTH SIDES\n x - 2.5 = +or- Square root of 12.5\n\nANSWER\n x = Sqrt 12.5 + 2.5   -or-\n x = 2.5 - Sqrt 12.5
965936
use pythagorean theorem.\n\na^2 + b^2 = c^2\n\npluggin in\n\na^2 + 4^2 = 11^2\n\na^2 + 16 = 121\n\nSolve for a\n\na^2 = 105\n\na = sqrt(105)\n\na = 10.247
1022403
Working with equation 2:\n\n5^2[2^(x-4) +5^(y-2)]=13*25\n5^2*2^(x-4) + 5^y = 325\n\nAdd this to 1st equation, yielding\n\n5^2*2^(x-4) + 2^(x+1) = 456\n2^(x-4)[5^2 + 2^5] = 456\n2^(x-4) = 456/(25+32)\n2^(x-4) = 8\nx-4 = 3\nx = 7\n\nUse 1st equation:\n\n2^8 - 5^y = 131\n5^y = 256 - 131 = 125\ny = 3
81890
it's not factored completely if you leave it as a^2-25 so the proper answer is 5(a-5)(a+5)
259349
Solve by substitution.\nSolve for y^2 in the second equation by subtracting x^2 and 2x from both sides.\nx^2 + y^2 + 2x - x^2 - 2x = 19 - x^2 - 2x\ny^2 = 19 - x^2 - 2x  Now substitute this into the first equation in place of y^2\nx^2 + 2(19 - x^2 - 2x) = 33  Distribute the 2\nx^2 + 38 - 2x^2 - 4x = 33  Combine x^2 and -2x^2 and subtract 33 from both sides\n-x^2 - 4x + 5 = 0  Multiply both sides by negative one (I just do this because it is easier to factor when the x^2 is positive)\nx^2 + 4x - 5 = 0  Now factor\nWe are looking for factors of -5 that add up 4.  5 and -1 work because 5*(-1) = -5 and 5 - 1 = 4\nTherefore x^2 + 4x -5 = 0 becomes\n(x+5)(x-1) = 0  Set each part equal to 0\nx+5 = 0 or x-1 = 0\nTherefore,\nx = -5 or x = 1\nNow substitute each of these values into the y^2 equation from above.\ny^2 = 19 - x^2 - 2x\nFor x = -5\ny^2 = 19 - (-5)^2 - 2(-5)\ny^2 = 19 - 25 + 10\ny^2 = 4  Take the square root of each side to get\ny = -2 or 2\n\nFor x = 1\ny^2 = 19 - (1)^2 - 2(1)\ny^2 = 19 - 1 - 2\nY^2 = 16  Take the square root of each side to get\ny = -4 or 4\n\nThus, the solutions are\n(-5,-2), (-5,2), (1,-4), (1,4)\n\nIt is always good to check your answers, especially on a test.  To check your answers just plug each one into each equation to see if you get a true statement.
678883
(2y)(2y)+14y=-24 \n=> 4y^2+14y+24=0\n=>2y^2 + 7y + 12 =0 As 2 is never 0\n=> y = [-7 +/- sqrt(7^2-4*2*12)]/4\n=> y = -7/4 +/- sqrt(49-96)/4\n=> y = -7/4 +/- sqrt(-47)/4\nSo the complex root of y are:\n-1.75+1.713914i and\n-1.75-1.713914i
652953
I got:\n1. 6x^2 - 36x +48\n2. 6x + 22
218422
You know that when x=0, y=50.  You want to know what x is when y = 25\n\n25 = 50*.88^x\n.5 = .88^x; take ln or log of both sides\nln(.5) = ln(.88^x); by the rules of logs, the x goes out in front\nln(.5) = xln(.88)\nx = ln(.5)/ln(.88) = approx 5.422 years.
785524
y = 3x-5\n\nx^2 +(3x-5)^2 = 25\nx^2 + 9x^2 -30x + 25 = 25\n10x^2 -30x = 0\n10x(x-3) = 0\n\nso x = 0 or x = 3\n\nWhen x = 0 y = -5\n\nWhen x = 3  y = 4
991546
you divide both sides by 25, you should have either a fraction or decimal, and then you should have x^2= 13/25, and then take the square root of both sides. you will have x=__.\n2(x-5)^2=3.............25 divide into 2 minus 3.........correct to solve for x\n\nx^2+4x+4=7\n.. You should get all numbers on one side to where it would be.. x^2+4x-3=0. then factor\n\ncompleting the square.\nx^2-6x-3=0\n\n2x^2-8x+5=0\n\nI never did understand completing the square. If you don't have to show your work on completing the square, then do it a way you do know. That's how I always did it. And then have your teacher sit down and show you.
758639
x^2-25x=50\nx^2-25x+625/4=50+625/4\n(x-25/2)^2=(200+825)/4(taking LCD and adding)\nx-25/2=+/-rt1025\nx=25/2+/-rt1025\n=25/2+/-5rt41
728357
3x^2y^2(2y+3y)/3x^2y^2\n=5y
983612
First swap the variables, then solve for R again.\n\nv = (21/2) - (1/64)R^2\nR^2 = -64v + 672\nR = ±√(-64v + 672)
845970
25x² - 20x = 1\n25x² - 20x - 1 = 0\n25x² - 20x - 1 + 5 = 0 + 5\n25x² - 20x + 4 = 5\n(5x - 2)² = 5\n5x - 2 = +/- √5\n5x = 2 +/- √5\nx = (2 +/- √5) / 5\nx = (2 + √5) / 5 and (2 - √5) / 5
725272
Take half of the coefficient of the x term, which comes to 7/2.\nSquare that, which comes to 49/4.\n(x + 7/2)^2 = x^2 +7x + 49/4