solve by simplying the equation square root of x - 2 = 1?
598233
just a little correction to taylor's answer.\nsqrt(x-2)=1, square it on both sides \n=> x-2 => square(1)=1\n=> x=3
797591
2...... 2....... 2 (2 is the square root)\nx  =17 .. +  ..10\ndo the same for the next one\nu can solve the equation ,I know u can
761784
Well, by inspection -2 is one root...\n\nP(-2) = (-2)^3 - 4(-2)^2 - 7(-2) + 10\nP(-2) = -8 - 16 + 14 + 10\nP(-2) = -24 + 24\nP(-2) = 0\n\nNow take your equation and divide by (x - (-2)).  In other words (x + 2):\n\n............ 1 -6 5\n1 2 ) 1 -4 -7 10\n....... 1 2\n....... ---------\n........... -6 -7\n........... -6 -12\n........... ----------\n............... 5 10\n............... 5 10\n............... ------\n.................... 0\n\nSo the result is x^2 - 6x + 5\n\nThis can be factored to (x - 5)(x - 1)\n\nThis results in (x + 2)(x - 5)(x - 1)\n\nAnd your roots are:\n-2, 5, 1\n\n(I don't know why I didn't notice 1 as a root... I usually check that first...)
826907
ill write down the formula for finding the root of a quadratic equation :  ax^2 + bx + c = 0\nsolution: x = {-b +- sqrt(b^2 - 4ac) } / 2a\n\nequation x^2=5x + 2 can be written as :\n\nx^2 - 5x - 2 = 0\nhere a = 1 ; b = -5 ; c = -2 ;\nsimply apply the formula to find the value of x .\n\nx ={ -(-5)+-sqrt(25 - 4*-2) } /  2\nx ={ 5 +- sqrt(24+8) } / 2\nx = { 5 +- sqrt(33) } / 2\n\nThe second equation :\n\n4x^2 + 2x - 3 = 0\nhere a = 4 ; b = 2 ; c = -3\n\nx = {-2 +- sqrt( 4 - 4*4*-3) } / 2*4\nx = {-2 +- sqrt(4+48) } / 8\nx = { -2 +- sqrt(52) } /8\n\nsimplify the term inside the square root ,  52 = 13*4\nso 4 gets square rooted and\n\nx = { -2 +- 2 sqrt(13) } /8\n\ntake 2 common from the numerator and denominator and cancel them.\n\nx = 2{ -1 +- sqrt(13) } / 4*2\nx = { -1 +- sqrt(13) } / 4\n\nthats your answer.\n\nhave a nice day :)
720929
Let me just work through the steps to solve the problem:\n\nsqrt(2x^2-1)=x   <---Original Equation\n2x^2-1=x         <---Square both sides\n2x^2-x-1=0       <---Subtract x\n\nNow, apply the quadradic equation:\nx=(-b+/-(sqrt(b^2-4ac))/(2a)\n\nFor this equation, the equation is:\nx=(-(-1)+/-(sqrt((-1^2)-4*2*(-1))/(2*(2))\n\nSimplified it is:\nx=(1+/-sqrt(1+8))/(4)\nx=(1+3)/4 or x=(1-3)/4\n\nSo, x=1 or x=-1/2
25136
1 = sqrt(1) = sqrt(-1*-1) = sqrt(-1)*sqrt(-1)=i*i = i^2 = -1\nSo 1 = -1.  Add 1 to both sides and 2 = 0.  Divide both sides by 2 and 1 = 0.
610821
Easy...\n0 = 0\n\nadding -2 on both sides\n\n-2 = -2\n-> -3 + 1 = -6 + 4\n-> -2*3/2 + 1 = -4*3/2 + 4\n\nadding 9/4 on both sides..\n\n-> 9/4 -2*3/2 + 1 =9/4 -4*3/2 + 4\n->\n(3/2 - 1)^2 = (3/2 - 2)^2\n\ntaking +ve sqr root\n3/2 - 1 = 3/2 - 2\n\n-> 1= 2\nadding 1 on both sides\n2 = 3\n-> 1 + 1 =3
891271
first into the second:\n\nx^2+(x-1)^2=25\n\nexpand:\n\nx^2+x^2-2x+1=25\n\nget 0 on the right side:\n\n2*x^2 - 2*x -24 = 0\n\na=2, b=-2, c=-24\n\nuse the root finding formula for quadratoic equation:\n\nx=(-b+/-sqrt(b*b-4*a*c))/(2*a)=(2+/-sqrt(4+4*2*24))/4=0.5 +/- 3.5\n\nyour solutions are: x = -3 and 4\n\ncheck:\n(-3)^2+(-4)^2 = 9 + 15 = 24, correct\n(4)^2+(3)^2 = 15 + 9 = 24, correct\n\nsubstitute to the first eqn: y = -4 for the 1st. x and 3 for the 2nd.
655260
2=-1=x\n2+1=x\nthere fore x =3
1172065
if it is a quadratic (ax^2 + bx + c) then put it into the quadratic equation. that automatically will give you 0, 1, or 2 real answers. the equation is: x = [-b +/- (square root of b^2 - 4*a*c)] / 2a\n\nif (b^2 - 4*a*c) is negative, you have no real solutions. if it is zero, you have one real solution. if it is positive, you have 2 real solutions.\n\nthen, if for example you get  x=4 and x= -2, your factors are (x-4) and (x+2)
346405
You have to solve the equation so that you can eliminate the sqrt by raising it to the second power .for instance,\n\nsqrt(2x+10)=5     you square it and get\n\n2x+10=25   easy, isnt it?\n\nbe careful if you have something like this\n\nsqrt(x+3)+sqrt(x-2)= 5\n\nhere you cannot do that you have to distribute one sqrt on each side of the equation.\n\nsqrt(x+3)=5-sqrt(x-2)\n\nnow you can square them
588927
easy.. -1 , -2 , 1,2,3.\n\n-1 x -2 =2\n2x2x3 = 12
369100
a(x^2)+2*(ax)+a=0\na(x^2+2x+1)\nLet a=0\nx^2+2x+1=0\n(x+1)^2=0\nsquare root of (x+1)^2=x+1 & square root of 0=0 \nSo,x+1=0\nx=0-1\nx= -1
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]
538847
For any quadratic of the form ax^2 + bx + c = 0, the roots are\n\nx = (-b +/- sqrt(b^2-4ac)) / 2a\n\nSqrt here is shorthand for "the square root of".
607013
Another Method.\n\n-1 = -1\n\n-1/1 = 1/-1\n\nSquare root both sides,\n\nSQR(-1/1) = SQR (1/-1)\n\nExpanding,\n\nSQR(-1)/SQR(1) = SQR(1)/SQR(-1)\n\nSQR(-1)/1 = 1/SQR(-1)\n\nCross Multiply,\n\nSQR(-1)*SQR(-1) = 1\n\n(SQR(-1))^2 = 1\n\n-1 = 1\n\nDivide by 2 on both sides,\n\n-1/2 = 1/2\n\nAdd 3/2 to both sides,\n\n-1/2 + 3/2 = 1/2 + 3/2\n\n1 = 2 (ans)
958505
x={-(-4) (+/-)√[(-4)^2 - 4(1)(-5)]}/(2(1))\n x= 4(+-)6/2\n x=5,   x=-1\n\nCheck:\n with x=5,  5^2-4(5)-5=25-20-5=0  OK\n with x=-1   (-1)^2 -4(-1) -5 = 1+4-5=0  OK