how to work do quadratic expressions and equations?
607601
I think you set the equation to zero then you have to factor it.  Go to the link below if you wanna know how to do it. OR look your your math book.
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~ ♥ ~
1079811
It *is* awfully hard to write algebra expressions on Yahoo Answers, isn't it? Oh well, we do what we can.\n\nAddressing your first question, let's write it out like a long division problem.  You'll have to imagine the horizontal bar above the larger polynomial, because I can't format it properly here:\n\nx^2 + 3x - 2  )  x^4 + x^3 - 5x^2 + 13x - 6\n\nOur first step is to look at the first term in each polynomial, and divide the smaller into the larger.  x^4 / x^2 = x^2, so the first term we write in our quotient (above the bar) is x^2.  You can write it directly above the x^4, if you wish, or above the 5x^2 term, your preference.\n\nNow, we're going to multiply that "partial quotient" by the entire divisor (x^2 + 3x - 2), which gives us x^4 + 3x^3 - 2x^2.  Write that "partial product" below the first three terms of the original dividend (x^4 + x^3 - 5x^2).  Our next step is to subtract the partial product: (x^4 + x^3 - 5x^2) - (x^4 + 3x^3 - 2x^2) = -2x^3 - 3x^2.  We then drop down the next term from the dividend, the 13x term, and repeat:\n\n-2x^3 / x^2 = -2x (add that term to our quotient)\n\n-2x * (x^2 + 3x - 2) = -2x^3 - 6x^2 + 4x\n\n(-2x^3 - 3x^2 + 13x) - (-2x^3 - 6x^2 + 4x) = 3x^2 + 9x\n\nFinally, we drop down the last term from the dividend, -6, and repeat once more.  I'm sorry that I can't write this out the way your math book can... I really hope Yahoo Answers creates some sort of way of inputting formulas, or at least monospacing text entry!\n\n3x^2 / x^2 = 3 (so we add a +3 term to our quotient)\n\n3 * (x^2 + 3x - 2) = 3x^2 + 9x - 6\n\n(3x^2 + 9x - 6) - (3x^2 + 9x - 6) = 0... no remainder!\n\nSo our final quotient is x^2 - 2x + 3, with no remainder.\n===\nNow, when doing the 2nd problem, you'll want to write the larger polynomial, the dividend, as x^4 - 3x^3 + 0x^2 + 6x - 5.  We incorporate that extra 0x^2 term so that we never find ourselves trying to subtract unlike terms.\n\nSorry there's no better way to write this out on Yahoo Answers.  Hope that helped anyway!
208740
EDIT: ** Baah. When I typed this answer I was first. Then at the end, I'm like the 3rd. Oh well..**\n\nThere are two ways. Which one is easier depends greatly on your thinking skills.\n\nYou can take the expression x2+7x+12 and assume it should break up in two factors:\n(x+ ? )(x+ ?)\nThen, search for two numbers that add up to the 7, but when multiplied equal 12.\nIt helps if you break down 12 into it's components, 12= 2*2*3. So, obviously, the answers could be:\n\n2 * (2*3) = 2 * 6 (but 2+6 is not =7)\n3 * (2*2) = 3 * 4, voila! 3+4 also = 7, so:\n\n(x+3)(x+4).\n\nThe second way could be just solving the cuadratic formula for second degree equations:\n\nb=(-b +/- SQRT(b2 - 4ac))/2a\n\nWhere ax2 + bx + c, so you'll have: a=1, b=7, c=12.\n\nReplace, then solve twice: Once for the + value of the SQRT, another for the - value of the SQRT. You should get something like:\n\nFirst run:\nx= -3\n\nSecond run:\nx= -4\n\nx=-3 is the same thing as (x+3)=0\nand \nx=-4 is also (x+4)=0\n\nNotice the two factors are precisely (x+3) and (x+4)?\n\nHope this helped..
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.
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
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
708644
Use the quadratic formula.\nx1 = 1\nx2 = -1/2
839760
The roots are 0 and -16/5\n\nTo solve any quadratic equation eg. ax^2+bx+c, you can apply formula x= (-b+-sqrt(b^2-4ac))/2a
781910
not possible to factor easily ... try the quadratic formula.
518727
Multiply the first equation by 2.\n\nSubtract the second equation from the new first equation.\n\nSolve for y.\n\nTake your value for y and substitute it into the original first equation.\n\nSolve for x.
825052
The first step would be to try to factor the equation, so you'd want to get all of the terms on the left side of the equation:\nx^2 + 4x -3 = 0\n\nSince this won't easily factor, you need to use a different method.  I'm sure someone will suggest the quadratic equation, which is fine _if you can remember it_.  I much prefer an alternate method, "completing the square," because it's intuitive enough that you can remember how to work out the process even if you forget the steps.\n\n1. Start by moving the constant to the right side of the equation:\nx^2 + 4x = 3\n\n2. Then divide the coefficient of x (in this case, 4) by two and square it, adding it to both sides of the equation:\nx^2 + 4x + 4 = 7\n\n3. Factor the left side of the equation to (x + a)^2, where 'a' will be the number you just squared:\n(x + 2)^2 = 7\n\n4. Take the square root of both sides, remembering that you'll have both a positive and a negative root:\nx + 2 = +/- (sqrt)7\n\n5. Move the constant on the left to the right, and you have your two solutions:\nx = -2 +/- (sqrt)7
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
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
797913
That's the quadratic formula. It is the solution to equations of the form\nax^2+bx+c=0
686655
No prob dude... take a deep breath, grab a cold drink, and concentrate on this... ready? ok....\n\nOften, the simplest way to solve "ax2 + bx + c = 0" for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy, or it doesn't factor at all, or you just don't feel like factoring. So, while factoring may not always be successful, the Quadratic Formula can always find the solution. \nThe Quadratic Formula uses the "a", "b", and "c" from "ax2 + bx + c", where "a", "b", and "c" are just numbers. The Formula is derived from the process of completing the square, and is formally stated as:\n\n For ax2 + bx + c = 0, the value of x is given by:\n   \nNote that, for the Formula to work, you must have "(quadratic) = 0". Note also that the "2a" at the bottom of the Formula is underneath everything above, not just the square root. And don't forget that it's a "2a" under there, not just a "2"! And make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back" on your test, and you'll mess yourself up. And remember that "b2" means "the square of ALL of b, including the sign", so don't leave b2 being negative, even if b is negative, because the square of a negative is a positive. In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the long run. Trust me on this!
66185
(-b +/- SQRT(b^2-4*a*c)) / 2a\n\nSo plugging in the values\n\nx = (10 + SQRT(100-4*16))/2\nand\nx = (10 - SQRT(100-4*16))/2\n\nx = 8\nand\nx = 2\n\nYou can find it written better in wikipedia:\n\nhttp://en.wikipedia.org/wiki/Quadratic_function\n\nGood thing we didn't have to get into complex numbers.... ;-)