how do you do quadratic equations?
406680
Do you mean 6x^2-3x-4?\n\nIf you do, we know that there are 3 ways of solving quadratics: factoring, completing the square, and quadratic formula. I would not recommend completing the square as it is long and involves tedious math word.\n\nAnother helpful hint: In standard form ax^2+bx+c=0, b^2-4ac is known as the discriminant. If it is negative, there are no solutions, and if it is not a perfect square, then you cannot factor.\n\nIn this quadratic, the discriminant would be (-3)^2-4(6)(-4) which equals 105. 105 is not a perfect square, so we cannot factor.\n\nQuadratic Formula: \n           -b +/-  sqrt (b^2-4ac)\nx= -------------------------------\n                2a\n\nsubstitute for a, b, and c
831674
1. replace x with -3, calculate and see if you get 0\n2. divide by (x+3) and solve the remaining quadratic equation
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
857303
Hmm. It could just be something like forgetting a parenthesis. Let's go through it step by step:\n\nThe quadratic formula is:\n\n[ -b +/- sqrt(b^2-4ac) ] / 2a\n\nBecause of the +/-, we will have two equations:\n\n[ -b + sqrt(b^2-4ac) ] / 2a\nand\n[ -b - sqrt(b^2-4ac) ] / 2a\n\nThe equation is x^2 + 3x - 84 = 0\na = 1\nb = 3\nc = (-84)\n\nSo we have:\n[ -3 + sqrt(3^2-4(1)(-84)) ] / 2(1)\nand\n[ -3 - sqrt(3^2-4(1)(-84)) ] / 2(1)\n\n[ -3 + sqrt(345) ] / 2\nand\n[ -3 - sqrt(345) ] / 2\n\n7.78709\nand\n-10.7871
734091
This particular quadratic equation will require the quadratic formula:\n\n-b +/- sqrt(b^2-4ac) all divided by 2a.\n\nHere, a=1, b=18, and c=49\n\n-18 +/- sqrt[(18^2 - 4(1)(49)] all divided by 2.\n\n18^2 - 4 x 49 = 128.....\n\nLet's do the top first:\n\n-18 +/- sqrt(128)\n\nWhat is sqrt(128)? You need to factor 128 = 64 x 2\nTherefore, sqrt(128) = sqrt(64 x 2) = 8 x sqrt(2).\n(we took 64 out of the root function, so comes out as the sqrt, which is 8.)\n\nSet it all up:\n\n-18 +/- 8sqrt(2) divided by 2:\n\n-9 +/- 4sqrt(2)\n\nAnswer: -9 - 4sqrt(2), -9 + 4sqrt(2).\n\nNot easy due to the answer not a set of integers.
795657
1. Move the constant to the right side of the equation:\n3x^2 - 8x = 6\n\n2. Divide out the x^2 coefficient:\nx^2 - 8/3x = 2\n\n3. Divide the coefficient of x (in this case, -8/3) by 2 and square it, then add it to both sides:\nx^2 - 8/3x + 16/9 = 34/9\n\n4. Factor the left side of the equation to (x + a)^2, where a is the number you just squared:\n(x - 4/3)^2 = 34/9\n\n5. Take the square root of both sides, remembering that this will produce both a positive and a negative root:\nx - 4/3 = +/-sqrt(34/9)\n\n6. Move the constant on the left to the right and simplify:\nx = 4/3 +/-sqrt(34/9) = 4/3 +/-(sqrt 34)/3 = (4 +/- sqrt 34)/3
217158
It's abc formula to solve the quadratic equation but the plus should be plus or minus
1039684
for x^2+5x-9=0  u can't do this the same way.9 * 1 = 9 but 9 + or - 1 not = 5 and 3 * 3 = 9 but 3 + or - 3 not = 5.\nso there is a formula u got to use here.\nit is  x = {- b + - sqrt(b^2 - 4ac)} / { 2a}\nso from the given quadratic equation,\na = 1,b = 5 and c = 9\nand then u substitute , u will get 2 values of x and that's the answer.remeber we have to use this formula if we are unable to factorise the way u did for the previous one.
737893
Take your quadratic equation.\n\nThe graph cuts the x axis when y = 0\n\nSo simply put your equation = 0, and then solve it either by using factorising or the quadratic formula:\n\n(-b plus or minus sqrt(b^2 +4ac)/2a)\n\nif I remember right, (could be minus 4ac)
905417
By dividing the equation by 2:\n\nx^2 + 3x +1 = 0\n\nThe quadratic formula is = [-b +/- sqrt(b^2 - 4*a*c)]/2a\n\nWhere a =1;     b = 3;       c = 1.\n\nNow, we plug in this values in the quadratic formula, and we'll obtain a value of x, using the minus sign, and other, using the plus sign, as follows:\n\nx1= [ - 3 + sqrt( 3^2  - 4*1*1)]/(2*1)\nx1= [ - 3 + sqrt(9 - 4)]/2\nx1= [ - 3 + sqrt(5)]/2\nx1= [ -3 + 2.236]/2\nx1= -0.382\n\nx2= [ - 3 - sqrt( 3^2  - 4*1*1)]/(2*1)\nx2= [ - 3 - sqrt(9 - 4)]/2\nx2= [ - 3 - sqrt(5)]/2\nx2= [ -3 - 2.236]/2\nx2= -2.618
36555
There are few ways of solving for the unknown in a quadratic equation.\n\nIn general, a quadratic equation is of the form:\n   a*x^2 + b*x + c = 0, a != 0    --- (1)\nDividing (1) by "a" through out, we get,\n   x^2 + (b/a)*x + (c/a) = 0  --- (2)\n\nOne way of solving is to really guess the solution. It works like this:\nAssume we have (somehow) factorised (2) into a form like:\n   (x - p)(x - q) = 0   --- (3)\n(3) can also be written as (by mutiplying term by term),\n   x^2 - (p+q)*x + p*q = 0   --- (4)\nComparing (2) and (4),\n   p + q = -b/a\n   p*q   = c/a\nSo, if the equation is written like in (2), the sum of roots is negative of x co-efficient, and the product is the constant. A "root" is a value of x that satisfies the given equation (1).\n\nFor e.g, given,\n   x^2 - 6*x + 5 = 0,\nthe sum of roots must be 6 and the product must be 5. So (we guess), the roots must be 1 and 5. Easy!\n\nBut "guessing" is not always possible :(\n\nLook at this equation:\n   x^2 - 6*x + 10 = 0\nThe roots of this are 3 + i and 3 - i. Not easy to guess right?\n\nA generic way of solving is called "completing the square". And it is a very clever piece of work! It works like this:\nEquation (2) again:\n   x^2 + (b/a)*x + (c/a) = 0\nThe aim is to get the equation into a form that looks like:\n   (x + something)^2 = somethingelse\nThen we can solve for x easily.\n\nLet us first get rid of the constant term to the other side:\n   x^2 + (b/a)*x = -(c/a)\nLet us re-write this again like:\n   x^2 + 2*(b/2a)*x = -(c/a) --- (6)\nLook at the left hand side (LHS) of the equation.. it is looking like: x^2 + 2*something*x. If only we have something^2, then we can put it in the form we want. So we'll add that term:\n   x^2 + 2*(b/2a)*x + (b/2a)^2 = -(c/a) + (b/2a)^2\nWe've got it! We now have,\n   ( x + (b/2a) )^2 = -(c/a) + (b/2a)^2\n\nWe can solve for x quite easily from here (left as exercise to the reader).\n\nThe "magic" equation (-b plus or minus sqare-root of etc...) is really derived by completing the square.
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!
839343
x^2 + 4x + 4 = 7 (subtrat 7 from both sides)\n\nx^2 + 4x -3 = 0 (use the quadratic equation formula)\n\n-b + or - the square root of (b^2 - 4ac) / a^2\n\n-4 + or - the square root of 16 - (4) (1) (-3) / 2\n\n-4 + or - the square root of 28 / 2\n\n-4 + or - (the square root of 4) (the square root of 7) / 2\n\n-4 + or - 2 (the square root of 7) / 2 (simplify by dividing 2 into  the -4 and 2 on the top to get rid of the 2 on the bottom)\n\nthat gives you x = - 2 + or - the square root of 7.
443949
Use quadratic formula and do your own homework.
1150325
The quadratic formula is:  [-b+/-((b^2-4ac))/(2a)] where a is the number befor x^2, b is the number before x and c is the integer.    Using this formula we get:\n\n1. x = [1+/- (1-(4)(6)(-12))^1/2/((2)(6))]\n    x = [1+/- 17]/(12)\n    x = 18/12 or -16/12\n    x = 1.5, x = -4/3\n\n2. x =  [6+/- (36-(4)(1)(13))^1/2/((2)(1))]\n    x = [6 +/- (16)^(1/2) i}(2) {since there is a negative sign inside the sq root we replace it by and positive sq root and imaginary number}.\n    x = [6 +/- 4i]/(2)\n   x = 3 + 2i, x = 3-2i
693558
On a graph with x and y-axes , if you plot out the values of y calculated from an equation involving x squared you draw out a parabola.  A quadratic equation is one in the form y = a.x-squared + b.x + c \n a can be positive or negative (not zero or you don't involve x-squared). If negative, the parabola is upside down (open end downward).\nb and c can be positive negative or zero.\nWhen you are asked to "solve" a quadratic, you are finding where the parabola cuts the x-axis, that is to say the values of x where the equation calculates out to y = 0