substitute u=x^2 into the equation x^-4 -32 = 4X6-2.. what is the quadratic equation?
458634
The equation becomes x^(-4)=32+22; or x^(-4)=54; x^4=1/54\n.: x^2=±1/√54; so that x=±1/∜54 are two real roots - that is regular numbers. There are two more roots, x=(±1/∜54).√(-1), which are special numbers called imaginary numbers.\n\nQuadratic eqn is an equ of the form ax^2+bx+c=0. where a, b, c are known, x is unknown. if you put x^(-2)=u, the eqn becomes u^2=54 which is a quadratic eqn. solve it and put back the value of u. What you get can be written as another pair of quadratic eqn. Solve this to get the value of x. The end.
439883
Use the following four steps to solve a quadratic equation, in this order:\n\nStep 1: Guess the answer\n\nIf x^2=0, can you guess the answer?\nIf x^2=1, you can guess, but remember that there may be two answers. (1 and -1)\n\nYou can tell if you are guessing right by plugging back into the original equation. If it worked, you're done.\n\nIf you can't guess in 30 seconds, you are done with Step 1. Go to Step 2.\n\n\nStep 2: Try factoring\n\nx^2 + 4x + 4 = (x+2)(x+2)\n\nRemember that the constant term of your factors has to multiply to the constant term of the quadratic. \n\nFor example, if you have x squared plus something plus 6, remember that your constant terms must be multiples of 6. \n\nTry:\n\n(x + 1)(x + 6)\n(x - 1)(x - 6)\n(x + 2)(x + 3)\n(x - 2)(x - 3)\n\nYou just ran out of factors of 6. If this didn't work, go to step 3.\n\n\nStep 3 -- Complete the Square\n\nOne important thing to remember about completing the square is that it does exactly the same thing as the quadratic equation. Pick either Step 3 or Step 4 as your last step. Guessing and factoring don't always work because you can't always guess and you can't always factor. But completing the square and using the quadratic formula will always find the answer if there is one.\n\nSuppose you have \n\nx^2 - 6x - 16 = 0\n\nHere's how to complete the square:\n\n#1. Move the constant (whatever it is) to the right side\n\nx^2 - 6x = 16\n\n#2. Find half of the coeficient of the x term and add the square of that number to both sides of the equation:\n\nThe x term here is -6x\nThe coefficient is -6\nHalf of -6 is -3 (Remember this number -3)\n-3 sqared is 9. Add 9 to both sides of the equation:\n\nx^2 - 6x + 9 = 16 + 9\n\n#3. If you did #2 correctly, then the left side is a perfect square and the constant term is -3 (remember -3?)\n\n(x - 3)^2 = 25\n\n(If you're not sure, multiply it out and see if you get what you had before.)\n\n#4. Take the square root of both sides. Remember, it could be negative or positive.\n\nx - 3 = 5\nx - 3 = -5\n\n#5. Solve your two equations\n\nx = 8\nx = -2\n\n#6. Put your answers back into the original equation and see if you got the right answer. \n\n\nStep 4. \n\nIf you did Step 3, you already have the answer, and you're already done. If you didn't do Step 3, use the quadratic formula to get the answer. You can find the quadratic formula here:\n\nhttp://en.wikipedia.org/wiki/Quadratic_formula\n\nMemorize the quadratic formula and never forget it.  Write it on your forhead until you have it memorized.
1007240
Let's start with an equation of the form: x^2 + bx + c = 0.\nThis means that x^2 + bx = -c.\n\nConsider the formula ( x + k )^2 = x^2 + 2k x + k^2.  Every perfect square is in this form.\nWe want to change x^2 + bx into a perfect square.  That means we need to determine what to add to put x^2 + bx into this form.\n\nBy equating coefficients:  bx = 2k x.  That means k = (b/2)\nThus k^2 = (b/2)^2.\n\nAdding (b/2)^2 to both sides of our earlier equation we arrive at:\n\nx^2 + b x + (b/2)^2 = -c + (b/2)^2\n\nWe added this to convert the left hand side into the form of a perfect square.  Thus we can factor it as a square.\n\n(x + b/2)^2 = -c + (b/2)^2\n\nTake the square root of both sides ( for now just assume that the right hand side is positive ).\n\n( x + b/2 ) = +/- sqrt( -c + (b/2)^2 )\n\nnow solve for x:\n\nx = -b/2 +/- sqrt( -c + (b/2)^2 )\n\nExample:\n\nSolve x^2 - 6x - 4 = 0\n\nPut into the proper form:  x^2 - 6x = 4\nDetermine k:  b = -6 so k = -3\nDetermine k^2:  k = -3 so k^2 = 9\nAdd k^2 to both sides:   x^2 - 6x + 9 = 13\nFactor left hand side:  (x - 3)^2 = 13    this is ALWAYS (x+k)^2\nSquare root:  x-3 = +/- sqrt( 13 )\nSolve for x:    x = 3 +/- sqrt( 13 )\n\n\nFor a more general equation:  a x^2 + b x + c = 0\nFirst divide through by a to get:  x^2 + (b/a) x + (c/a) = 0\nThis is now in the earlier form so use that method will solve it.\n\nGood Luck!
164954
x^3 – 3x^2 + 4x – 12\n= x^2(x – 3) + 4(x – 3)\n= (x-3)(x^2 +4)
730173
x = (-b +/- sqrt(b^2 - 4ac))/2a\n\nwhere a= 1, b= -6 , c=8\n\nx1 = (6 + sqrt(36 - 32))/2 = 4\nx2 = (6 - sqrt4)/2 = 2
219130
I could not get your question. We all know that the quadratic equations are those equation which contain a square with the unknown (x). Quadratic equation will always have two roots. But, where did you get the "Four"? Would you explain a bit more.
612720
-2 = 1/(1+x)-3/(x-1)\nmultiply both sides by (x^2-1)=(x+1)*(x-1):\n\n-2(x^2-1)= x-1 -3(x+1) \n\nrearrange\n2x^2 - 2x - 6 = 0\ndivide by 2\nx^2 - x -3 =0\napply formula for quadratic equations\nx=(1 +/- sqrt(1+4*3))/2\n = 0.5 +/- sqrt(13)/2\n\n"+/-" means plus or minus (there are two solutions)\n"^2" means squared
929526
4x^2 - 16 = 0\n4(x^2 - 4) = 0\nx^2 - 4 = 0\n(x - 2)(x + 2) = 0\n\nANS : x = 2 or -2
962913
because it is quartic equation it can be factored into 2 quadratic with real coefficients. let us try if real\nx^2+ax+b and x^2+cx+d as coefficient of x^4 is 1\n\nmultiply and equate coefficent on both sides\n\na+c = 6 (x^3)\nbd = 66(constant)\nb+d+ac = 17(x^2)\nad+bc = 36(x)\n\nif they are rational they are integer\nfrom bd =66 let b= 6 and d= 11(we can chose one)\nthen a+c = 6\nac =0\n11a+6c = 36\nso a= 0 and c = 6\nequation is consistat\nso we get (x^2+6)(x^2+6x+11)\n\nthese cannot be factored without going to irrational and complex
502865
x^2-2x=24\n=> x^2-2x-24=0\n=> (x-6)*(x+4)=0\n=> x=6 and x=-4
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
160067
When you substitute with y=x^(1/6), you're assuming :\nI) x>0 (which should be put as a starting condition)\nII) y>0 . This is to be considered after solving the 2nd degree equation, as you get a fake solution y= -3 that you must reject as non-acceptable.\n\nThe only acceptable solution is x= 1/64
945218
[(x^2)^2]^2\n\n(x^2 -4x +4)^2\n\nmultiply      x^2  (x^2 -4x +4)^  = x^4 -4x^3 +4x^2\n\nmultiply     -4x (x^2 -4x +4)^    = -4x^3 +16x^2 -16x\n\nmultiply     4  (x^2 -4x +4)^      =  4x^2 -16x +16\n__________________________________________\n\nthe answer should be                 x^4 -8x^3 +24x^2 -32x +16\n\ni hope that helped
952497
Ok, I assume you know the quadratic equation, but just for convenience I'm going to write it out:  x = (-b +/- root(b^2 - 4ac))/2a.  \n\n1) x = (5 +/- root(25 + 24))/4.  (Remember, you have to subtract the 3 from each side to get the equation to equal 0, so c is -3.)  That simplifies to x = (5 +/- 7)/4--->x = 12/4 or -2/4--->x = 3 or -1/2.\n\n2) x = (2 +/- root(4 - 12))/6.  Are you sure that wasn't supposed to be a negative 3x^2 or minus 1?  If it were, the term inside the square root would be 16, a positive perfect square.  As it is, it's -8, whose square root is 2root(2)i.  That seems pretty complex (no pun intended) compared to the other one.  \n\nI'm going to solve that one assuming that it should have ended in -1 rather than 1.  Then x = (2 +/- root(4 + 12))/6--->x = (2 +/- 4)/6--->x = -2/6 or 6/6--->x = -1/3 or 1.
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
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
749124
x^2 + 4x + 4=7  (first put it in the form ax^2 + bx + c = 0, by subtracting 7 from both sides of the equation)\n\nx^2 + 4x - 3 = 0   (this is called a quadratic equation)\n\nNow substitute a =1, b = 4, c = -3 in the following formula\n\nx equals -b plus or minus the square root of (b^2 - 4ac) all over 2a\n\nBecause you are solving for x^2 you expect to get 2 answers.  The plus or minus in the middle of the formula gets you 2 answers as you solve the formula for both plus and minus the square root of (b^2 minus 4ac).\n\n[If you're going to have to do this in an exam, it's probably worth learning the formula off by heart.  It's boring to have to do so but you'll probably have more confidence in the exam if you see an quadratic equation and already know the formula.]\n\nGood luck for the working out - you can do this bit!