Solve using the square root property:?
490168
5x^2 =100\nx^2 = 100/5\nx^2 = 20\nx = + or - sqr root of 20\n\n(x+8)^2 = 81\nx+8 = 9\nx=1
670230
x^2+4x+Y^2-8Y+16=0\nX^2+4x+Y^2-8Y=-16\nX^2+4x+4+Y^2-8Y+16=-16+16+4\nX^2+4x+4+Y^2-8Y+16=4\n(X+2)^2+(Y-4)^2=4\nR=Square root 4\nR=2
141530
Steps:\n1) Divide by 2\nx^2=32\n2)Find square root\nx=sq.rt.32\n3)Simplify Radical\nx=sq.rt.(2)(16)\nANSWER:\nx=4+sq.rt.2\n\n*=times
832801
Completing the square involves adding constants to both sides of your equation so you have a form:\n\n(x - a)^2 = b\n\nOnce you get this form, you can take the square root of both sides:\n\nx - a = sqrt(b)\n\nx = a + sqrt(b) \n\nSince a square root can be either positive or negative, it is just as correct to write:\n\nx = a - sqrt(b)\n\nNow, back to the original "perfect square"\n\n(x - a)^2 = b\n\nExpanding this out:\n\nx^2 - 2*a + a^2 = b\n\nMoving b to the other side:\n\nx^2 - 2*a + a^2 - b = 0\n\nYour equation is 2*x^2 - x - 2 = 0\n\nForst thing is to get plain x^2, so divide by 2:\n\nx^2 - (1/2) * x - 1 = 0\n\nNow from the form above, a = 1/.4 because 2*a = (1/2)*x\nNext, look at the constant term: \n \n-1 = a^2 - b\n-1 = (1/4)^2 - b\n\nSolving for b: b = 17/16\n\nSo your perfect square equation is:\n\n(x - 1/4)^2 = 17/16\n\nSolving for x:\n\nx - 1/4 = sqrt(17/16)\n\nx = 1/4 + sqrt(17/16)\nx = 1/4 - sqrt(17/16)\n\nSimplifying slightly:\n\nx = (1 + sqrt(17))/4\nx = (1 - sqrt(17))/4
205798
Always remember:\n"The Product of square roots of 2 numbers is equal to the square root of their product"\nIn symbols, \n(√a)(√b) = √(ab) and\n√(ab) = (√a)(√b)\nso,\n2√12 = 2√(4*3)\nUsing our 'Property',\n2√12 = 2(√4)(√3) but √4 = 2, so\n2√12 = 2(2)(√3)\n2√12 = 4√3\n\nHope this helps^_^
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
665576
x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4
319669
did you mean the square root of x-1 = 3?\nin that case it will be 2:\nadd positive 1 to both sides:\n   x-1+1= 3+1\nthen solve the remaining equation:\n   x=4\nget the square root of 4:\n   x=2
764942
√(4x + 1) +3 = 0\nSo √(4x + 1) = -3 (which means that 4x + 1 cannot be real as √N  ≥ 0)\nSo no solution\n\nIf you want to solve: \n\nFnd an x such that one of the square roots of (4x + 1) is - 3,\n\nthen 4x + 1 = 9\nx = 2
770289
It's the difference between two perfect squares.  Take the square root of 16 and x squared.  Then a minus sign and the square root of 25.  Put parentheses around all of that. Write it again, and change the minus sign to a plus sign.  You will set each group in parentheses equal to zero.  There will be 2 equations.  Solve for x.  X will have 2 solutions.
800980
1. Move the constant to the right side of the equation:\n4x^2 + 2x = 1\n\n2. Divide out the x^2 coefficient:\nx^2 + 1/2x = 1/4\n\n3. Divide the coefficient of x (in this case, 1/16) by 2 and square it, then add it to both sides:\nx^2 + 1/2x + 1/16 = 5/16\n\n4. Factor the left side of the equation to (x + a)^2, where a is the number you just squared:\n(x + 1/4)^2 = 5/16\n\n5. Take the square root of both sides, remembering that this will produce both a positive and a negative root:\nx + 1/4 = +/-sqrt(5/16)\n\n6. Move the constant on the left to the right and simplify:\nx = -1/4 +/-sqrt(5/16) = -1/4 +/-(sqrt 5)/4 = (-1 +/- sqrt 5)/4
533051
16.3707055437449\n\nYou can use the free online scientific calculator at the link below to solve square root problems.
830100
'x= rcost\ny=rsint\n\nand r= sqrt(x^2+y^2)\nt=arctan(y/x)\nso\nif x=2, and y=2, then r= sqrqt(4+4)=sqrt(8)\nand t=arctan(2/2) = arctan(1) = pi/4
1187142
maybe something like 7^(1/3) over 7^(1/2)\n\nthat's the same as the cube root of 7 over the square root of 7.\nTo solve it, you would just subtract the exponents and you'd get 1 over 7^(1/6)\n\nI hope that is what you are looking for.
781653
Eliminate the y to get\n2-x^2=2x+4\nx^2+2x+2=0\nusing quadratic formula\n(-2+-sqrt(4-4*2))/2\nwe see there are no real roots: the two functions do not intersect\nx= -1+-i
358881
http://www.nist.gov/dads/HTML/squareRoot.html\nhttp://www.math.com/students/calculators/source/square-root.htm\n\nI hope it will help
574453
answer are below                        suareroot ==sqrt\n1. x^2-8x+10=0                       \n   (x-4)^2-16+10=0\n   (x-4)^2=6\n   (x-4)= sqrt(6)\n   x=4+sqrt(6) or x=4-sqrt(6)\n2. x^2+20x-1=0\n   (x+10)^2-100-1=0\n   (x+10)^2=101\n   (x+10)=sqrt(101)\n    x= -10+sqrt(101) or  x=-10-sqrt(101)\n3. 3x^2-6x+5=0\n   we can`t solve using square method\n4. x^2-16x-10-7=0\n   (x-8)^2-64-10-7=0\n   (x-8)^2=81\n    (x-8)=+or-9\n    x=17or-1\nthank u