Algerbra solving equation qn 4?
91753
factorize\nx^2 + x = x(x+1)\n\nfactorize\nx^2 - 1 = .... (you should know)\n\n2 = 2 (!!!)\nTake the lcm of the three denominators (each different factor (=parenthesys) once and with the highest power)\n\nIt is 2x(x+1)(...)   (fill the dots)\n\n.....\n\nnow if you can't go on, you need to open your book again and study, or pay more attention at school at your math classes, and ask immediately for further explanation to your teacher when you feel you've not understood.\n\nGood work and good study
921629
You can easily solve this by realizing (4m - 3n)(4m + 3n) has a pattern (a + b)(a - b), which equals to a^2 - b^2\n\nSo,\n(4m - 3n)(4m + 3n) = (4m)^2 - (3n)^2\n= 16m^2 - 9n^2
295616
D (the discriminant) = b^2 - 4ac = 25 - 4*(-2)*0 = 25\nSince D>0  there are two real solutions.
800980
1. Move the constant to the right side of the equation:\n4x^2 + 2x = 1\n\n2. Divide out the x^2 coefficient:\nx^2 + 1/2x = 1/4\n\n3. Divide the coefficient of x (in this case, 1/16) by 2 and square it, then add it to both sides:\nx^2 + 1/2x + 1/16 = 5/16\n\n4. Factor the left side of the equation to (x + a)^2, where a is the number you just squared:\n(x + 1/4)^2 = 5/16\n\n5. Take the square root of both sides, remembering that this will produce both a positive and a negative root:\nx + 1/4 = +/-sqrt(5/16)\n\n6. Move the constant on the left to the right and simplify:\nx = -1/4 +/-sqrt(5/16) = -1/4 +/-(sqrt 5)/4 = (-1 +/- sqrt 5)/4
986257
If the equation is X^4 - 12x^2 -64 =0, there is.\n\nUse u = x^2, now your equation will look like this:\n\nu^2 - 12u - 64 =0\nFind u1 and u2 using the quadratic formula.\n\nThe solutions are u1 = 16, u2 = -4\n\nNow you go back to x, first deal with u1, then u2.  Because u = x^2 each u solution will give you two x solutions.\n\n\nu = x^2 = 16\nx = +/- SQRT(16)\nx1 = 4\nx2 = -4\n\nx2 = -4.  This is not possible with real numbers. so it will give you two imaginary numbers. Do the same as before\n\nx = +/- SQRT(-4) = +/- 2*SQRT(-1) = +/- 2i\n\nx3 = i\nx4 = -i
508029
The equation is:\n(x+a+b+c)^2 = 2x^2+(x+a)^2-(x+b)^2-(x+c)^2\nA simple rearrangement yields\n(x+a+b+c)^2 - x^2 = (x+a)^2-(x+b)^2 + x^2-(x+c)^2\nFactorize using the identity a^2-b^2=(a+b)(a-b)\n(x+a+b+c+x)(x+a+b+c-x) = (x+a+x+b)(x+a-x-b) + (x+x+c)(x-x-c)\nwhich can be written as\n(2x+a+b+c)(a+b+c) = (2x+a+b)(a-b)-c(2x+c)\n==> 2x(a+b+c)+(a+b+c)^2 = 2x(a-b)+(a-b)(a+b)-2xc-c^2\n==> 2x(a+b+c-a+b+c) = (a+b+c)^2 + a^2-b^2-c^2\n==> 4x(b+c) = a^2+b^2+c^2+2ab+2bc+2ca+a^2-b^2-c^2\n            = 2(a^2+ab+bc+ca) = 2(a+b)(c+a)\n==> x = (a+b)(c+a)/[2*(b+c)]
636051
x=4\n\nplease solve...\n\n8-(1) :-) ;-(
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
952054
To find the equation, just substitute the 2 roots to the "Quadramagic formula" (...)\nx² - (p + q)x + pq = 0\nwhere p and q are the roots.\n\nHence, p = 7 + i and q = 7 - i.  We substitute:\nx² - [(7 + i) + (7 - i)]x + (7 + i)(7 - i) = 0\n\nWe add and multiply\nx² - (14)x + (7² - i²) = 0\n\nSince i² = -1, then\nx² - 14x + 49 - (-1) = 0\n\nthen,\nx² - 14x + 49 + 1 = 0\n\nTherefore, the quadratic equation is\nx² - 14x + 50 = 0\n\nTherefore, the correct answer is 1).\n\n^_^
1053403
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
530596
Given\nax^2 + bx + c = 0\nDivide all terms by a \nx^2 + (b / a) x + c / a = 0 \nSubtract c / a from both sides \nx^2 + (b / a) x + c / a - c / a = - c / a \nand simplify \nx^2 + (b / a) x = - c / a \nAdd (b / 2a)^2 to both sides \nx^2 + (b / a) x + (b / 2a)^2 = - c / a + (b / 2a)^2 \nto complete the square \n[ x + (b / 2a) ]^2 = - c / a + (b / 2a)^2 \nGroup the two terms on the right side of the equation \n[ x + (b / a) ]^2 = [ b^2 - 4a c ] / ( 4a^2 ) \nSolve by taking the square root \nx + (b / a) = ± sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } \nSolve for x to obtain two solutions \nx = - b / a ± sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } \nThe term sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } may be written \nsqrt { [ b^2 - 4a c ] / ( 4a^2 ) } = sqrt(b^2 - 4a c) / 2 | a | \nSince 2 | a | = 2a when a > 0 and 2 | a | = -2a when a < 0, the two solutions to the quadratic equation may be written \nx = [ -b + sqrt( b^2 - 4a c ) ] / 2 a \nx = [ -b - sqrt( b^2 - 4a c ) ] / 2 a \nThe term b^2 - 4a c which is under the square root in both solutions is called the discriminant of the quadratic equation. It can be used to determine the number and nature of the solutions of the quadratic equation. 3 cases are possible \ncase 1: If b 2 - 4a c > 0 , the equation has 2 solutions. \ncase 2: If b 2 - 4a c = 0 , the equation has one solutions of mutliplicity 2. \ncase 3: If b 2 - 4a c < 0 , the equation has 2 imaginary solutions\nps. I did this as quick as possible, hope everything copied correctly.
145305
Use cubic equation \ny=ax3+bx2+cx+d\n\nSince you have four points data you can get four linear equations with four variables which is very easy to solve, get variables (a,b,c,d) and plug the values to cross check and see u will get answer. Use guass jordan or elimination method.
750949
I'm not sure of the exact equations, but I know the side that's 4 times longer should be 4 + x, and the one side should be x.  Then using the Pathagorem thm., x^2+ ?^2= (4+ x)^2\n\nI'm just not sure how to find the third side.\n\nHope that helps :/
740569
1) L = 4W\nA = LW = 20\nSubstitute L = 4W to get 20 = 4W x W\n20 = 4W^2, divide both sides by 4 to get\n5 = W^2, take square root of both sides to get W\nW = √5\n\n20 = L x W\n20 = L x √5, divide both sides by √5 to get L\nL = 20/√5 or, when rationalized, 4√5\nFinal answer:\nL = 4√5 and W = √5\n\n----\n\n2) L = W + 4\nA = 6 = L x W, substitute L = W + 4 to get\n6 = (W + 4) x W\n6 = W^2 + 4W, substract 6 from both sides to get\n0 = W^2 + 4W - 6\nThis is not easily factored, so you'd have to use the quadratic equation to solve for W. Once you have W, just plug that in to L = W +4 to get the L value. Good luck!
676814
First equation: sqrt[x] + y = a\nor sqrt[x] = a - y\nSquaring both sides\nx = (a - y)^2__________(1)\n\nSecond equation: x + sqrt[y] = b\nor sqrt[y] = b - x\nSquaring both sides\ny = (b - x)^2___________(2)\n\nSubstituting 'x' from eqn. (1) in eqn. (2)\ny = [b - (a - y)^2]^2\n  = [b - a^2 + 2*a*y - y^2]^2\n\nSo it is clear that you have to solve a 4-th order equation.\nIt is really tough.
504744
4. \nquad 1 = (x,y)\nquad 2 = (-x,y)\nquad 3 = (-x,-y)\nquad 4 = (x,-y)
412542
x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a