How do I put a quadratic equation in transformational form?
82007
Just type in quadratic equation and maybe roots in a serch engine. You will have your pick of dozens of sites.
914755
your equation must be like x^2-3x-6 and your trying to factor this\n\n2 numbers say a and b must multiply to make -6 and add to make -3 so a*b=-3 and a+b=-6\n but i can't think on any that fits this criteria so you must use the quadratic equation\n\nx= [-b +-squareroot(b^2-4ac)]/2a\nwhere a=1 b=-3 and c=-6\nx= 3+-squareroot(9-4(1)(-6))/2\n\nx=(3+squareroot(33))/2 and x= 3-squareroot(33)/2\n\nin decimals x=4.3723 and x=-1.3723
147771
You have to write as a product of two terms.\nThis is always possible, as a consequence of an important theorem. \nIn general, if you have the expresion a*x^2 + b*x + c, where a,b,c are constants and x is a variable, here is the way you can put this expression in a product form.\nFirst calculate b^2-4*a*c (the so-called determinant). If this number is negative the expression can't be put in the product form (well at this stage).\nThen calculate x1=(2*a-sqrt(determinant))/2 and x2=(2*a+sqrt(determinant))/2. The result will be:\na*x^2 + b*x + c= (x-x1)*(x-x2).\nIn your case the determinant is 256 and sqrt(256)=16. then x1=-7 and x2=9 and that's it.
1057648
You always can bring a quadratic equation\nax2 + bx + c = 0  in the form x2 + 2px + q = 0\nby dividing with a and p = b/(2a) while q = c/a\n\nx2 + 2px + q = 0\nx2 + 2px + p2 + q = p2\n(x + p)^2 = p2 - q\nx + p = +V(p2 - q) or x + p =  -V(p2 - q) \nx  = -p +V(p2 - q) or x = -p  - V(p2 - q)\n\nWhen you replace p and q you find the famous abc formula.\nSatisfied?
21539
If you have a system of equations with two variables of the form ax + by = c, dx + ey = f, where a,b,c,d,e,f are the coefficients, and x and y are variables, you can set up the equation as \n\n[a b][x] = [c]\n[d e][y] = [f]\n\n(ignore the fact there are two equals signs there, they're only there for formatting reasons)\n\nwhere the square brackets are a crude attempt on by behalf to 'draw' matrices and column vectors.\n\nTo solve the system you need to find the inverse of the matrix of the right hand side (assuming it's invertible to begin with), and "right multiply" (that is multiply from the right of each side of the equation) both sides of the equation.\n\nThat should give you:\n\n[1 0][x] = [ inverse of matrix ][c]\n[0 1][y] = [ ........................ ][f] \n\nThen it's just a matter of performing the calculation.\n\n(sorry for the crappy formatting of the post)
265248
the general form of a quadratic equation ax^2 + bx + c =0\n\ntherefore there is two solution\nx=[-b ± sqrt (b^2-4ac)]/(2a)\n\nx^2+2x+2=0\n\na=1 b=2 c=2\n\nx1=[-2+sqrt(2^2-4*1*2)]/2 =(-2+2i)/2 = -1+i\nx2=[-2-sqrt(2^2-4*1*2)]/2 =(-2-2i)/2 = -1-i\n\nwhere i is the square root of -1 and it is imaginary\n\nthe solution is X={-1+i , -1-i}\n\nThose solution means that this equation in the from y=x^2+2x+2 has no intersection with y=0 since no real solution for x^2+2x+2=0
1160760
what you have is something like\na(b+c)\n\nthe solution is \nab+ac\n\n(x^3)(2x^-2)+(x^3)(4x)\nnow to multiply x^a * x^b you get x^(a+b)\n\n2x^((3+(-2))+4x^(3+1)\n\n2x+4x^4
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.
736851
I got a different answer than you.  Let's do this together:\n\nRecall that for an equation ax² + bx + c = 0, that the quadratic formula is:\n\nx = [ -b ± √ (b² - 4ac) ] ÷ 2a\n\nSo we want to find the zeros of x² + 10x + 23\n\nSo, x = [ -b ± √ (b² - 4ac) ] ÷ 2a\n\n= [ -10 ± √ (10² - 4*1*23) ] ÷ 2*1\n\n= [ -10 ± √ (100 - 92) ] ÷ 2\n\n= [ -10 ± √8 ] ÷ 2\n\n= [ -10 ± 2√2 ] ÷ 2\n\n= -5 ±√2\n\nHope that helped.\n\n~ ♥ ~
291823
the quadratic formula is a derivation of the equation \n\nax^2+bx+c=0 \n\nit is: (-b+/-sqrt(b^2-4ac))/2a\n\nit gets the answer out of any and all quadratic trinomials equal to zero...\n\nhere is a site that shows the derivation process:\nhttp://www.purplemath.com/modules/sqrquad.htm#formula
210912
Put the left side coefficients into matrix [A],\n[6 2]\n[5 6]\nand the right side values into matrix [B]\n[-16]\n[..4]\nthen tell the calculator to multiply [A] inverse * [B]\nthe result is the solution matrix\n[x]\n[y]\n\nnote, the inverse key looks like x^-1.  \n\nsolve the second system the same way.  coefficient matrix will be 3x3, answer matrix 3x1.
893823
Use the quadratic equation \n\nt = -b + or - sqr(b^2 - 4ac)/2a\n\nyour equation is already in the form:\n\nat^2 + bt + c = 0\n\njust plug in the variables and solve for t twice
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.
708644
Use the quadratic formula.\nx1 = 1\nx2 = -1/2
768295
Using an equation is the lazy, mechanical way to do it. If you want to understand what's going on, factor it if possible.\n\nIf you *really* want to know what's going on, derive the quadratic formula yourself. It isn't as hard as it looks, and it can be done on one page.
371573
If i is a root, you may express it into x = i, a root of x^3 + i = 0.\nsince x = i, x - i = 0\nTherefore, x^3 + i has a root x - i. to get the other 2 factors, divide.\n(x^3 + i)/(x - i) = 0\nDivide it using long division, and you'll get\nx^2 + ix - 1 = 0\nActually, \nx^3 + i factors as (x - i)(x^2 + ix - 1) \ntherefore, \nx^2 + ix - 1 = 0\nuse the quadratic formula:\na = 1\nb = i\nc = -1\nx = [-b +or- sqrt(b^2 - 4ac)]/2a\nx = [-i +or- sqrt(i^2 - 4(1)(-1))]/2(1)\nx = [-i +or- sqrt(-1 + 4)/2\nx = (-i +or- sqrt -3)/2\nx = (-i +or- i sqrt3)/2\nTherefore, the other roots are\nx = (-i + i sqrt3)/2\nx = (-i - i sqrt3)/2\n\n^_^
750403
I have to go teach my class in about 4 minutes, but when I return, if no one has done it, I can show you how.  It requires factoring out a specific value that makes the coefficients "a" and "b" turn into two values that satisfy x^2 + y^2 = 1.  In other words, I know it starts out by factoring out sqrt(a^2 + b^2) from \n\na sin(Bx)+b cos(Bx)\n\n\nto get \n\n\nsqrt(a^2 + b^2) [ a/sqrt(a^2 + b^2) * sin (Bx) + b/sqrt(a^2 + b^2) * cos(Bx)]\n\n\nEDIT: Okay here we go.  And the "orthogonal rotating vectors" proof below is usually over most people's heads, or at least, most people who would ask this question on Yahoo answers.\n\n\nFirst, you factor out SQRT(a^2+b^2) as I mentioned above.  For the sake of simplicity, I will let A = SQRT(a^2+b^2).  So factor A out of  \n\na sin(Bx)+b cos(Bx)\n\nto get\n\nA [a/A sin(Bx) + b/A cos(Bx)]\n\n\n\n\nSince (a/A)^2 + (b/A)^2 = 1 (you might need to actually verify this for your purposes), there exists some angle C such that\n\ncos (C) = a/A and sin (C) = b/A \n\n(also, you should point out that both a/A and b/A have absolute values less than one, so that C is real and not complex with a nonzero imaginary part).\n\nSo make the substitutions to get\n\nA [cos(C) sin(Bx) +  sin(C) cos(Bx)].\n\n\n\n\n\nAccording to the formula \n\nsin(y+z) = sin y cos z + cos y sin z, \n\nthe expression in the brackets must equal sin(Bx + C).  So now the whole expression becomes\n\nA sin (Bx + C).\n\n\n\n\n\nSo in conclusion, \n\na sin(Bx)+b cos(Bx) = Asin(Bx+C),\n\nwhere A = SQRT(a^2+b^2), and C = arccos(a/A).  \n\n\n\n\nThere is a better representation for C, however.  \n\nSince cos C = a/A and sin C = b/A, you have\n\ntan C = (sin C)/(cos C) = (b/A)/(a/A) = b/a.\n\nThus, tan C = b/a, and C = arctan (b/a).  This definition of C does not rely on the creation of a third variable (A).\n\n\n\n\n\nHope this helps!