let a=2\n\nmultiply both sides by a\n\n(a)^2 = 2a\n\nSubtract 4 from both sides\n\n(a)^2 - 4 = 2a - 4\n\nor\n\n(a+2) x (a-2) = 2( a-2)\n\n(a-2) is common on both sides\n\nhence \n \na+2 = 2\n\nNow a =2 (as above)\n\nnow\n2+2 = 2\n\nor\n4 = 2\n\nor \n\n2 = 1\n\nHOW IT CAN BE POSSIBLE???\n\nPlease tell me the reason....!!!
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the process u follow is correct, but actually it is wrong. fundamentally wat u did is ok, but technically it is wrong. because this becomes a quadratic equation which has more than 1 root. so this equation has more than 1 answers for which it is true.
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The top 2 terms have a in common; factor it out by Distributive Property: a(a+4)\n\nThe bottom 2 terms have b in common, do the same: b(a+4)\n\nNow as long as a doesn't equal -4, the (a+4) on top cancels the (a+4) on bottom, leaving a/b
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Prove that it really is possible! I really would like to see it.\n\nAssume x = y\n\nThen\n\nx + y = y + y  (Add y to both sides of the equation)\n\nx - y + y = y  (Subtract y from both sides)\n\n(x-y+y)/(x-y) = y/(x-y)\n\nSimple Regrouping:\n(x-y)/(x-y) + y/(x-y) = y/(x-y)\n\nWe all see that (x-y)/(x-y) = 1 we can also\nAdd 1 + 1 to both sides \n\n1 + y/(x-y) + 1 + 1  = y/(x-y) + 1 + 1\n\nHey, actually bot sides have a common term of y/(x-y) so let's just get rid of them by subtract them from both sides\n\n1 + 2 = 1 + 1\n\nHey, what do you know.\n3 = 1 + 1\n\nOk what did I do wrong guys. :p
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Use factoring.\nx2-2x-8=0\n(x-4)(x+2)=0\nso\nx-4=0 and x+2=0\nsolve\nx=4 and x=-2
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if they differ by 4i they are a + 2i and a- 2i\n\nsum = 2a = -2\n\na= -1\n\nroots are -1+ 2i and -1- 2i\n\nso (x+1)^2 = (2i)^2\nor x^2 +2x + 1 = -4\nor x^2+2x + 5 =0\n\nthe form is a+2i and a- 2i\n\nbecause if it is a+b and a-b the diff is 2b which is 4i so b = 2i
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(a^2 -2ab + b^2) = (a-b)(a-b) = (a-b)^2\nYou cannot cancel anything, so \n(a^2 -2ab + b^2)/(a+b) does not simplify.\n\nYou may wish to write it as (a-b)^2 /(a+b)
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ill give it a shot - im not sure , anyways i got 2 answers..\n\n(1st)\n\n(3-x)x+2 = 2 (x^2-2)\n\n3x-x^2+6-2x = 2x^2-4\n\nx-x^2+6 = 2x^2-4\n\nx-1/x+6 = 2/x-4\n\nx-1+6x = 2-4x\n\nx+4x+6x = 2+1\n\n11x = 3\n\nx = 3/11\n\n\n(2nd possible answer)\n\n\n(3-x)x+2 = 2 (x^2-2)\n\n3x-x^2+6-2x = 2x^2-4\n\nx-x^2+6 = 2x^2-4\n\nx-1/x+6 = 2/x-4\n\nx-1+6/x = 2-4/x\n\n-5 = -2/x\n\n-5x = -2\n\nx = -2+5\n\nx = 3
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First equation is A + 2 = 2(B - 2)\nSecond equation is B + 2 = A - 2\n\nSolve the second equation for A and substitute it into the first:  B + 4 = A\n\nB + 4 + 2 = 2B - 4\nB + 6 = 2B - 4\n10 = B\n\nThen figure out A
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First question:\nTake the sq. root of both sides:  x+2=+/-(sqrt5)\nThen subtract 2:  \n                               x = -2 +/-(sqrt 5)\n\nSecond prob:  subtract 5 first:\n\n(a+2)^2 = -3\nTake the square root of both sides:  (a+2) = +/-(sqrt(-3)\n    \nsqrt (-3) = i*sqrt 3\n\na+2 = +/-(i*sqrt 3)\n\nThen subtract 2: final answer:  -2 +/-(i*sqrt3)
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If squares of two numbers are equal the two numbers need not be equal always i.e. there is no such axiom or rule. \n\nLet us explain. \nSuppose x^2=y^2. \nThen x^2-y^2=0 [Transposition rule]\nor (x-y)*(x+y)=0 [Factorising]\n\nAs product of two numbers is zero so at least one of them must be zero [since anything multiplied by zero is zero] so\neither (x-y)=0 or (x+y)=0.\n\nWhen x-y=0, x=y but when x+y=0, x=-y.\nNow 'x' and 'y' are such numbers that there squares are equal but that not imply x = y because x = -y is also possible.\n\nSo your step "rooting both sides" is wrong leading to the wrong result.\n\nUNDERSTAND?
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Yes, there is a general form\n\nLet A the first number and B the second\n\nTo have A+B = AxB the general form\n\nA=1+x and B=1+1/x   where x is any number else zero\n\nx=1 A=2 B=2       2x2 = 2+2\nx=2 A=3 B=1.5     3x1.5 = 3+1.5\nx=3 A=4 B=1.333.. 4x1.333 = 4+1.3333\nx=4 A=5 B=1.25    5x1.25 = 5+1.25\n\nand so on\n\nThe proove\n\nA=1+x  B=1+1/x\n\nA+B = (1+x) + (1+1/x) = 2 + x + 1/x\nAB = (1+x) (1+1/x) = 1 + x + 1/x +1 = 2 + x +1/x\n\nSO A+B =AB
520147
Let x = "one number". \nThen, the "another number" is 2x.\n\nWhen the larger number is decreased by 10,\n\n2x - 10\n\nThe result is 2 greater than the smaller number\n\nx + 2\n\nTherefore,\n\n2x - 10 = x + 2, and\nx = 12.\n\nSo the two numbers are x and 2x, or 12 and 24.\n\nLet's put it back into the problem.\n\nOne number is twice another.  24 is twice 12.\nWhen the larger number is decreased by 10 (24 - 10 = 14), the result is 2 greater than the smaller number (2+12 = 14).  Therefore that's the answer!\n\nHappy solving!
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its simple distributive law.\nwhat has happened is they have taken (a-2) common out of both the terms .\neg: cx-x = x(c-1)\nthe same thing has happened here with x=(a-2) and c=a^2\nso, in step 2, we have: a^2(a-2)-(a-2)\ntaking (a-2) common outside; you end up with (a-2)*((a^2)-1)\nwhihc is [3]\nthen as you know, (a^2)-1 is same as (a^2)-(1^2) and then we apply the algebraic identity (a^2)-(b^2)=(a+b)*(a-b) to get (a-2)(a+1)(a-1) which is [4]
126028
You have to find a common denominator\nFirst, factor each denominator\nx^2 - 16 = (x+4)(x-4)\nx^2 + 8x + 16 = (x+4)(x+4) = (x+4)^2\nThe common denominator would be (x-4)(x+4)^2\nThe first term:\n(3x+1)/{(x+4)(x-4)} is missing an (x+4) so multiply the top and bottom by x+4\n{3x+1)(x+4)/{(x-4)(x+4)^2}\n= (3x^2 + 13x +4)/{(x-4)(x+4)^2}\nThe second term\n(3x+5)/(x+4)^2 is missing an x-4 so multiply the top and bottom by x-4\n(3x+5)(x-4)/{(x-4)(x+4)^2}\n= (3x^2 - 7x - 20)/{(x-4)(x+4)^2}\nNow that there is a common denominator, you can just subtract the numerators\n{3x^2 + 13x +4 - (3x^2 - 7x - 20)}/{(x-4)(x+4)^2}\n= (3x^2 + 13x + 4 - 3x^2 + 7x + 20)/{(x-4)(x+4)^2}\nAnswer: (20x + 24)/{(x-4)(x+4)^2}\nYou can factor a 4 from the numerator but nothing will cancel with it, so this is an acceptable final answer.
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Wow, what an interesting question. I'm surprised no one has solved it yet, but that may be a consequence of the fact that no one has understood it yet. Clearly, schools need to focus more on reading comprehension ^_^.\n\nTo answer your question - not only can we prove that there exist integers a and b satisfying your assumptions, but we can even show a method of finding them. The proof is long, so sit back, relax, grab your favorite drink, and prepare to read.\n\nFirst, consider that any solutions of these equations must obey the quadratic formula: x=(-b±√(b²-4ac))/2a. Substituting the coefficients of the  actual equations in, we find that the solutions of the first equation are (-p±√(p²-4q))/2 and the solutions of the second equation are (-p±√(p²+4q))/2. Let us consider only the greater of the solutions in each pair - that is, replace each ± with a +. The difference between these solutions is then (-p+√(p²-4q))/2 - (-p+√(p²+4q))/2. Call this number K. We can then perform some simple algebra as follows:\n\n(-p+√(p²-4q))/2 - (-p+√(p²+4q))/2=K\nmultiply both sides by 2 and simplify\n√(p²-4q)-√(p²+4q)=2K\nsquare both sides\n(p²-4q) - 2√((p²-4q)(p²+4q)) + (p²+4q)=4K²\nsimplify\n2p²-2√(p^4-16q²)=4K²\ndivide both sides by 2\np²-√(p^4-16q²)=2K²\nsubtract 2K² and add √(p^4-16q²) to both sides\np²-2K²=√(p^4-16q²)\nsquare both sides\np^4-4K²p²+4K^4=p^4-16q²\nmultiply both sides by -1 and simplify\n16q²=4K²p²-4K^4\ndivide both sides by 4\n4q²=K²p²-K^4\nadd K^4 to both sides\nK^4+4q²=K²p²\ndivide both sides by K² and rearrange\np²=K²+4q²/K²\n\nThus, if we let a=K and b=2q/K, then p²=a²+b² as required by the problem. All that remains is to show that a and b are nonzero integers.\n\nLemma 1: K is an integer\nProof: All solutions of both quadratic equations are integers. K is the difference of two of these solutions - that is, K is the difference of two integers. Therefore, K is an integer.\n\nLemma 2: K is nonzero\nProof: K is the difference between a solution of the first equation and a solution of the second equation. Suppose K=0. Then these equations have a common solution - that is, for some x, x²+px+q=0=x²+px-q. This would imply that q=-q. This can only be true if q=0. But q≠0. Therefore, the equations do not have a common solution and K≠0.\n\nLemma 3: 2q/K is an integer\nProof: Earlier we showed that √(p²-4q)-√(p²+4q)=2K. We can perform some simple algebra as follows:\n\n√(p²-4q)-√(p²+4q)=2K\nmultiply both sides by (√(p²-4q)+√(p²+4q)) and simplify\n2K(√(p²-4q)+√(p²+4q)) = (p²-4q)-(p²+4q)\nsimplify further and rearrange:\n-8q=2K(√(p²-4q)+√(p²+4q))\ndivide both sides by -4:\n2q=-K(√(p²-4q)+√(p²+4q))/2\ndivide both sides by K:\n2q/K=-(√(p²-4q)+√(p²+4q))/2\n\nNote that 2K=√(p²-4q)-√(p²+4q) Adding 2√(p²+4q) to both sides of that identity gives 2K+2√(p²+4q) = √(p²-4q)+√(p²+4q). We can then substitute to get:\n\n2q/K=-(2K+2√(p²+4q))/2\nSimplifying at last yields:\n2q/K=-K-√(p²+4q)\n\nWe already know K is an integer, so if √(p²+4q) is also an integer, then 2q/K is the difference of two integers and therefore an integer. Now we know that the solutions of the second equation are (-p+√(p²+4q))/2 and (-p-√(p²+4q))/2. The difference between them is √(p²+4q). However, since both of the solutions to the second equation are integers, their difference must be an integer. Therefore √(p²+4q) is an integer, and by extension, 2q/K is an integer. Q.E.D\n\nLemma 4: 2q/K is nonzero\nWe know 2q/k=-K-√(p²+4q). Suppose 2q/K=0. Then -K-√(p²+4q)=0 From this we would get:\n\n-K-√(p²+4q)=0\nadd K to both sides and rearrange:\nK=-√(p²+4q)\nmultiply both sides by 2\n2K=-2√(p²+4q)\nremembering 2K=√(p²-4q)-√(p²+4q), we substitute:\n√(p²-4q)-√(p²+4q)=-2√(p²+4q)\nadd √(p²+4q) to both sides and rearrange:\n-√(p²+4q)=√(p²-4q)\n\nHowever we specified in the beginning t
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To prove equality, we must start with the left hand side of the equation (LHS) and end up with the thing on the right hand side (RHS).\n\nThen, we must start with the thing on the RHS and then end up with thing on the LHS.\n\nFirst, let a and b be real numbers.\n\nCase 1: Assume a = b. Then,\n\n(LHS) = 0 = (a+b)*(a-b) = (a+b)*0 = (a+a)*0 = 2a*0 = 0\n\n(RHS) a^2-b^2 = a^2-a^2=0. Thus, LHS=RHS.\n\nRHS = a^2-b^2 = a^2-a^2 = 0.\n\nLHS = (a+b)*(a-b) = (a+b)*0 = (a+a)*0 = 2a*0 = 0. Thus, RHS=LHS.\n\nSince LHS=RHS and RHS=LHS, (a+b)*(a-b)=a^2-b^2.\n\n\nCase 2: Assume a does not equal b. Then,\n\nLHS = (a+b)*(a-b) = a^2 -ab +ba -b^2\n   \n                  = a^2-ab+ab-b^2\n\n                  = a^2-b^2\n\n                  = RHS.\n\nRHS = a^2-b^2  \n   \n    = (a+b)(a-b) [Difference of two squares]\n\n    = LHS.\n\n\nSince RHS = LHS and LHS = RHS, (a+b)*(a-b)=a^2-b^2.
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Are you missing a plus sign and parentheses around a+b?  If so, then\n(a^2 -2ab + b^2) = (a-b)(a-b) = (a-b)^2\nYou cannot cancel anything, so \n(a^2 -2ab + b^2)/(a+b) does not simplify.\nYou may wish to write it as (a-b)^2 /(a+b)\n\nIf you are not missing a plus sign or parentheses, then\na^2 - 2ab b^2 = a^2 - 2ab^3\nYou can factor out an a.\n= a(a-b^3)  This does not factor, so you cannot simplify.\nDivide this by a and you get a - b^3\nNow add b\nAnswer: a - 2b^3 + b