what is the square root of (9x^4 + 4x^2 + 1)?
963328
Uh, probably nothin'.  Did you mean 9x^4 + 6x^2 + 1?
156166
that's easy.\nD=b^2-4ac\nx1= [-b-sqrt(D)]/2a\nx2= [-b+sqrt(D)]/2a\n-------------------\na,b,c are the coeficients (ax^2+bx+c=0) so a=1, b=3, c=-5;\nsqrt is square root
216863
((4 * 4) / 4) - 4 + 4 = 4
800533
Well dear Kristi \n\nthe answer is \n( 9r - 2s )^2 \n{ as you know ( a - b )^2 = a^2 -2ab+ b^2 \n81r^2 - 36rs + 4s^2\na = 9r , b = 2s , so \na^2 = (9r)^2 = 81r^2 \nb^2 = (2s)^2 = 4s^2 \n-2ab = (-2)(9r)(2s) = -36rs  } \n\nGood Luck Darling
694690
determine the square of 6 - sqrt[(x-2)^2 + y^2]\n\n-->\n\n36 + [(x-2)^2 + y^2] - 12*sqrt[(x-2)^2 + y^2]\n= 40 + x^2 + 2x + y^2 - 12*sqrt[(x-2)^2 + y^2]
1128882
4*4+4+Sqrt(4) = 22, \n\nthis is 4*4 = 16\nplus 4 = 20\nplus the square root of 4, which is 2 = 22\n\nusing only the number four!
552978
sqrt(14404) \n= sqrt(4*3601)\n= 2*sqrt(3601)
835718
9^x - 4^x = (9 - 4)[9^(x - 1) + 9^(x - 2) * 4 + 9^(x - 3) * 4^2 + ..... + 9 ^(x - k ) * 4^(k - 1)  + ..... + 9^2 * 4^(x - 3) + 9 * 4^(x - 2) + 4^(x - 1)]\n= 5*[9^(x - 1) + 9^(x - 2) * 4 + 9^(x - 3) * 4^2 + ..... + 9 ^(x - k ) * 4^(k - 1)  + ..... + 9^2 * 4^(x - 3) + 9 * 4^(x - 2) + 4^(x - 1)]
839343
x^2 + 4x + 4 = 7 (subtrat 7 from both sides)\n\nx^2 + 4x -3 = 0 (use the quadratic equation formula)\n\n-b + or - the square root of (b^2 - 4ac) / a^2\n\n-4 + or - the square root of 16 - (4) (1) (-3) / 2\n\n-4 + or - the square root of 28 / 2\n\n-4 + or - (the square root of 4) (the square root of 7) / 2\n\n-4 + or - 2 (the square root of 7) / 2 (simplify by dividing 2 into  the -4 and 2 on the top to get rid of the 2 on the bottom)\n\nthat gives you x = - 2 + or - the square root of 7.
779215
(-4x^4)² = (-4x^4) * (-4 x^4)\n= (- 4) *( -4) * (x^4) * (x^4)\n= +16 * x^(4 + 4)\n= 16x^8
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
219026
x^3+4x+1\n\n(3x^4+2x^3+12x^2+11x+2)=\n=(3x+2)*(x^3+4x+1)
598163
The answer is\n\n(1/2)(x sqrt(4x^2+1)+ (1/2)ln(2x+ sqrt(4x^2+1)))\n\n\nheres the formula: int(sqrt(a^2+x^2))=\n(x/2)(sqrt(x^2+a^2))+\n(a^2/2)(ln(x+sqrt(x^2+a^2)))\n\nmight look complicated but take it down in a piece of paper and it will be pretty easy..
243752
Well, the definition of a square root has been explained pretty well.  The square root of 4 is 2.  sqrt(9)=3.  sqrt(2)=sqrt(2).  This is in simplist form becuase you cannot write sqrt(2) in a finite decimal expansion.\nHow to Simplify:\n\nConsider sqrt(99).\n99=11*9.\nso, sqrt(99)=sqrt(11*9)=sqrt(11)*sqrt(9)=sqrt(11)*3\nThe steps to follow are as described:\n1) try to break down the number being considered into its prime factors.  In this case: 11,3,3\n2)  look at those factors.  Are any of them perfect squares?  In our case, 9 is 3*3.\n3)  Use this property: sqrt(a*b)=sqrt(a)*sqrt(b) to break thye original number up into an irreducible number and a perfect square if possible.\n\nSome notes:\nThis is not possible if the number does not reduced to perfect squares.  For example: 143=11*13 neither of which are perfect squares so we are stuck here and sqrt(143) is in simplist form.  Hope this helps.
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
595518
1 + 4.9t^2 = 6t\n4.9t^2 - 6t + 1 = 0\n\nusing the quadratic formula\n\nt = (-b ± sqrt(b^2 - 4ac))/2a\nt = (-(-6) ± sqrt((-6)^2 - 4(4.9)(1)))/2(4.9)\nt = (6 ± sqrt(36 - 19.6))/9.8\nt = (6 ± sqrt(16.4))/9.8\nt = (6 ± sqrt(4 * 4.1))/9.8\nt = (6 ± 2sqrt(4.1))/9.8\nt = (3 ± sqrt(4.1))/4.9\nt = about 1.02548 or .199011\n\n-----------------------------------\n\n1 - 4.9t^2 = 6t\n4.9t^2 + 6t - 1 = 0\n\nt = (-6 ± sqrt(36 + 4(4.9)))/9.8\nt = (-6 ± sqrt(36 + 16.4))/9.8\nt = (-6 ± sqrt(52.4))/9.8\nt = (-6 ± sqrt(4 * 13.1))/9.8\nt = (-6 ± 2sqrt(13.1))/9.8\nt = (-3 ± sqrt(13.1))/4.9\nt = about -1.37312 or .148627
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