Design, code and test a program that solves a quadratic equation .Your program?
451055
hehe .. the above answer is funny!!\n\nthe above program simply squares a number and returns.\n\nLet the eqn be of form ax^2 + bx + c =0\n\nprocess is dumb actually.. first find delta (delta = b^2 - (4 *a*c))\n\nroots of equation are : [-(b) + sqrt(delta) ]/(4*a*c) AND [-(b) - sqrt(delta) ]/(2*a)\n\n\nok.. here's a program to solve a quadratic equation :\n\nits in java ..\n\n\nimport java.io.*;\nclass quadra\n{\npublic static void main(String args[])throws\n\nIOException\n{\nint a,b,c;\n\ndouble delta,root1,root2;\nSystem.out.println(" Enter coefficient of x^2 ");\n\nBufferedReader br = new BufferedReader(new\n\nInputStreamReader(System.in));\n\na = Integer.parseInt(br.readLine() );\n\nSystem.out.println(" Enter coefficient of x ");\n\n\nb = Integer.parseInt(br.readLine() );\n\nSystem.out.println(" Enter coefficient of\n\nconstant");\n\nc = Integer.parseInt(br.readLine() );\n\ndelta = (b*b) - (4*a*c);\n\nroot1 = ((-(b)) + Math.sqrt(delta)) / (2*a);\nroot2 = ((-(b)) - Math.sqrt(delta)) / (2*a);\n\nSystem.out.println("roots are"+ root1 +"and"+\n\nroot2);\n\n}\n}\n\n\nhope i have answered ur question\n; )\n\nif u want program in C\nmail me: anand_87123@yahoo.com\nsubject line : Y! answers - programming
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.
891987
The first step is to apply the Rational Roots Test to the polynomial to get a list of values that might possibly be solutions to the polynomial equation. You can follow this up with Descartes' Rule of Signs, if you like, to narrow down which possible zeroes might be best to check. Of course, if you've got a graphing calculator, it's a good idea to do a quick graph, since x-intercepts of the graph are the same as zeroes of the equation. Seeing where the graph looks like it crosses the axis can quickly narrow down your list of possible zeroes.\n\nOnce you've found a value you want to test, you use synthetic division to see if you can get a zero remainder. If you get a zero remainder, you've not only found a zero, but you've also reduced your polynomial by one degree. Remember that synthetic division is, among other things, division, so checking if x = a is a solution is the same as dividing out the linear factor x – a. This means that you should not return to the original polynomial for your next computation (for finding the other zeroes); you should instead work with the output of the synthetic division. It's smaller, so it's easier to work with.\n\nYou should expect to get some complicated solutions (that is, solutions containing square roots or complex numbers, or both). You will find these zeroes by applying the Quadratic Formula for the final quadratic factor.\n\nFind all the zeroes of the following polynomial: 2x5 + 3x4 – 30x3 – 57x2 – 2x + 24 \nFirst, I'll apply the Rational Roots Test--\n\nWait. Actually, the first thing I'll do is check to see if x = 1 or x = –1 is a root, because these are the simplest roots to test for. This isn't an "official" first step, but it can often be a timesaver, because you can just look at the powers and the numbers. When x = 1, the polynomial evaluates as 2 + 3 – 30 – 57 – 2 + 24 = –60, so x = 1 isn't a root. But when x = –1, I get –2 + 3 + 30 – 57 + 2 + 24 = 0, so x = –1  is a root, and I can do the synthetic division with x = –1 right now.\n\n\n\nThis leaves me with the smaller polynomial 2x4 + x3 – 31x2 – 26x + 24. (Since I've divided out the factor x + 1, I've reduced the degree of the polynomial by 1. That's how I know this is a degree-four polynomial.) Now I'll apply the Rational Roots Test to get a list of values to look at:
952526
x^2-5x-2=0\n\nuse the quadratic formula\n\n[5  +/- sqrt(25-4*1*-2)]/2\n\n[5  +/-   sqrt(25+ 8)]/2\n\nI distributed the 1/2...easier to type\n\n5/2  +/-sqrt33/2
809000
There are many ways to approach this question the easiest being if you have a graphing calculator just plug in a few points ie. (23,9200) (24, 9216)<look below for explanation> into stats and then use the stats operation quadreg and plug that formula into your y values and look at the table for the biggest y or you can look at the graph and use the maxium feature.\n\nif you don't have a graph calc then you need to grab a few points ... I'll use (23,400) (24, 384) (25, 368) (26,352) and (27, 336)... now multiply the x and y for each and then put that in the y value for each x.... ie. (23,9200) (24, 9216) (25, 9200) (26, 9152) (27, 9072)\n\nmake make a small t-chart (with 2 extra columns the first d and then d') with your first two as your new x then y points.... with the first extra column(d) take the difference of each y(for the first entry you will not be able to have an answer) ie. 16; -16; -48; -80; now in the second extra column(d') take the difference of the d column and multiply it by -1 ... -32, -32, -32\n\nnow you ready to start making your quadratic... take d'(32) and take it to the -2 power<I may be wrong here and it might be the -1 power, but you can just test and come back if wrong> then place that number in front of x^2 (I will refer to this number as a)\n\nok now take this formula ax^2 +bx +K(this is the constant but in this problem you have none) now plug in a value you have for x from your second points and solve for b now this is your quadratic, you should test it on a point or two, but use this to create a huge t-chart of different points and look for the when the value gets smaller, and the point before it gets smaller is your point\n\nwe actually ran into the answer while doing the problem which is common in math books but the point (24, 9216) or when you have 24 trees is the maximum yield
806483
Write down all the possible pairs which multiply to 70\nand pick out that pair which sums to 17.\nOr, try to solve the system\nx + y = 17\nxy = 70.\nSo y = x/70,\nx + 70/x = 17.\nNow multiply each term by x and solve the\nresulting quadratic.\nGood luck!
180174
Confirm, if you insert the values for x and y the equations work out with the solutions.
1076564
What do you want to solve for? x? y? or do you want an ordered pair (x,y) that works (integers maybe?)
685803
I was just looking for this this morning for my class!\n\nTry a search using words like:\nquadratic practical example application\n\nFirst of all, realize that a quadratic equation deals with "squares" both literally and in reality. So, think about a problem where the shape of a square is involved. \n\n1. For example, what size of piece of cardboard would you need to make a box of some desired volume? \n\n2. Anytime you have a movement in one direction with a resistance (wind, wheel friction, etc.) in any other direction, a quadratic equation might be useful.\n\n3. Whenever something is rising or falling and a gravity has an effect on its movement, a quadratic equation applies and can be useful.\n\n4. Also, any situation where you have matching of every member of a group with all the others of the group, like the classic handshaking problem or a sports round robin.\n\nGood luck with the assignment and remember that mathematics is the only set of rules that everything must obey and if you do it right and use it only for the purposes it is intended for, it will never lead you astray. Mathematics can theoretically be used for any real world situation but the trick is knowing when mathematics is the fastest or best way. Other ways are estimation, trial and error, and asking someone with experience.
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
797913
That's the quadratic formula. It is the solution to equations of the form\nax^2+bx+c=0
832512
either square root or erauqs
756992
Don't you just love the quadratic equation?\n\nThe first thing you got to do is get the given equation into the correct form, and in order to do that you have to combine all like terms. Since there are no like terms yet you have to do some math first.\n\n-2.0^2 = 4.0because neg * neg = pos\n\nNow you have 2 like terms 4.0 and 0.5 combine to get 4.5. Your eqution should look like this:\n\nf(t)=10t + 4.5 since you need at^2 + bt^2 + c just insert t^2. Now your equation will look like this:\n\nf(t)=t^2 + 10t + 4.5\n\nAnd just in case you need the quadratic equation:\n\nt= -b + or - sqr(b^2 - 4ac) / 2a\n\noh yeah and a = 1
803501
1.  add 1 toboth sides, then multiply by 3\n\n(x+4)^2=18  now take the square root of both sides\n\nx+4= +/- 3sqrt2\nx=-4 +/- 3sqrt 2\n\nyou also could use the quadratic formula\n\n\n2.  multiply num and denom by 3 - 2 sqrt3\n\n12-8sqrt3 +2sqrt3 -2sqrt6  all over -3\n\n\n3.  mult num and denom by 2 - sqrt 12 and get \n\n12-6sqrt12 all over -8\n\n12-12sqrt2  all over -8\n\n3-3sqrt2 all over -2
537199
got to math.com, click on calculators and click on solve equations.....it'll do the math for ya
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
156840
Look up "Algebra Solved 2005" or "algebrator", these two programs not only solve problems, but they give step by step solutions for understanding. they have tutorials and practice test. I went from knowing none to fairly advanced in a few weeks.