solve using the square root property   5x^2=100?
98037
2. Take the square root of both sides, thereby negating the ^2 and giving 10. Then solve for 5x=10.
117989
do it like grade 5 - 1 see - \n[X-(1/5X)^2] sqrt\n    (1/5X)^2 \n     1/25 (2X) \n[X-1/25(2X)]sqrt\nX=4 , understand
918825
Yes. When 'c' is given, find the value of c^2.  Then, find out the possible numbers, whose sum = that number. Find out the root of all these numbers, till you arrive a perfect square root( if possible ).\n\nEg: when given, c = 10, find c^2. Here, it is 100. \n   therefore, a few possibilities are : 50 + 50 ( none are perfect squares)\n                                                         25 + 75 ( only 1 of them is a perfect square)\n                                                         36 + 64 ( both are perfect squares)\n\n Therefore, the answers being looked for are 6 and 8.
747917
add -100 x^2\n0=4x^4-98.5x^2\nx^2(4x^2-98.5)=0\nx^2=0 or 4x^2-98.5=0\n4x^2-98.5=0\nx^2=98.5/4\nx=+/-rt(98.5)/2\nso x=0 or +/-(rt98.5)/2
968475
(2x sqrt(2x)) (3 sqrt(5x))\n= (2x*3) (sqrt(2x*5x))\n= (6x) (sqrt(10x^2))\n= (6x) sqrt(x^2) sqrt(10)\n= (6x)(x) sqrt(10)\n= (6x^2) sqrt(10)
1152199
It doesn't have a square root. A square root has to be an exact number (no decimals). Because it means that it is whatever that number is squared. ex: 100=10^2(10 squared). If it's a decimal number that goes on forever, it's impossible to square it because it's kinda infinite.
978895
Can you use a square root symbol?  If so, sqrt(9) + 9 - 9
412542
x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a
733063
5x^2-35x+3x-21\n5x^2-32x-21
746280
c^2 = 3^2 + 4^2\nc^2 = 9 + 16 = 25\nc = square root (25) = 5
629659
x^4 + 200 =102x^2\n   x^4 -102x^2 + 200 = 0        Now split the middle term\n   x^4 -100x^2 - 2x^2 + 200 = 0 Factor the first two and last two terms\n   x^2(x^2-100) -2(x^2-100) = 0\n   (x^2-2)(x^2-100) = 0\nEither (x^2-2) = 0  or (x^2-100) = 0\n        x^2 = 2     or  x^2 = 100\n        x= sqrt(2), -sqrt(2)    or x= +10, -10
675662
Step 1:  Isolate the radical term.\nSqrt(5x)=x-4\n\nStep 2:  Square both sides.\n5x=(x-4)^2\n\nStep 3:  Expand, and combine like terms.\n5x=x^2-8x+16\nx^2-13x+16=0\n\nStep 4:  Use the quadratic formula.\nx=[-b+/-Sqrt(b^2-4ac)]/2a\n=[13+/-Sqrt(169-64)]/2\n=(13+Sqrt105)/2 and (13-Sqrt105)/2 which is approx.\n=11.6235 and 1.3765\n\nStep 5:  Check the answers.\nThe first one works in the original equation.\nThe second one does not.  It is an extraneous solution.\n\nThe only solution is (13+Sqrt105)/2.\n\nNOTE:  Purdue Engineer was close, but he was mistaken about which answer is extraneous.
66185
(-b +/- SQRT(b^2-4*a*c)) / 2a\n\nSo plugging in the values\n\nx = (10 + SQRT(100-4*16))/2\nand\nx = (10 - SQRT(100-4*16))/2\n\nx = 8\nand\nx = 2\n\nYou can find it written better in wikipedia:\n\nhttp://en.wikipedia.org/wiki/Quadratic_function\n\nGood thing we didn't have to get into complex numbers.... ;-)
1095338
2^5/2=square root(2^5)=square root(32)=5.66\n\n2^6/2=square root(2^6)=square root(64)=8
1157065
(x+5)(2x^2+4x+2)\nx(2x^2+4x+2)+5(2x^2+4x+2)\n2x^3+4x^2+2x+10x^2+20x+10\n2x^3+14x^2+22x+10
800483
Let f(x)= y = √{100 - x²}.\n\ndy =½ * -2x/√{100 - x²}. dx\n= -x//√{100 - x²}. dx\n\nSo δy ≈ -x//√{100 - x²}.δx\n\nf(6) = √{100 - x²}\n= √{100 - 6²}\n= 8\n\nf(6.2) = f (6 + 0.2)\n≈f (6) + -6/8 * 0.2\n= 8 - 0.150\n= 7.850\n\nCheck\n\nf(6.2) = √{100 - (6.2)²}\n= √{100 - 38.44}\n= √{61.56}\n= 7.846 ......
775585
-5+-sqr(25-4(2)(-9))/4\n-5+-sqr(97)/4\n1.21 and -3.71