How do i model this problem with a quadratic equation and solve?
445638
l=2w, and length times width is area, so...\nl*w=A...use l=2w, w=w, A=5000\n2w*w=5000, or\n2w^2=5000...divide by two and solve for w!
101802
I don't think there is a unique solution for it. You have more unknown variables than the equations.
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!
881174
use formulas for cos(2x) and sin(2x) depending on cos(x) and sin (x) and then use the fact that\nsin(2x) = sqrt(1-cos(2x)^2).\nyou will get a quadratic equation that you solve with quadratic formula
831674
1. replace x with -3, calculate and see if you get 0\n2. divide by (x+3) and solve the remaining quadratic equation
851212
hmm im not quite in college yet, but i hope to be help.\ndo you know the values for f, h, v, and a?\nif you do, id substitute it in.\nthen square both sides. \nand then solve.
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
826907
ill write down the formula for finding the root of a quadratic equation :  ax^2 + bx + c = 0\nsolution: x = {-b +- sqrt(b^2 - 4ac) } / 2a\n\nequation x^2=5x + 2 can be written as :\n\nx^2 - 5x - 2 = 0\nhere a = 1 ; b = -5 ; c = -2 ;\nsimply apply the formula to find the value of x .\n\nx ={ -(-5)+-sqrt(25 - 4*-2) } /  2\nx ={ 5 +- sqrt(24+8) } / 2\nx = { 5 +- sqrt(33) } / 2\n\nThe second equation :\n\n4x^2 + 2x - 3 = 0\nhere a = 4 ; b = 2 ; c = -3\n\nx = {-2 +- sqrt( 4 - 4*4*-3) } / 2*4\nx = {-2 +- sqrt(4+48) } / 8\nx = { -2 +- sqrt(52) } /8\n\nsimplify the term inside the square root ,  52 = 13*4\nso 4 gets square rooted and\n\nx = { -2 +- 2 sqrt(13) } /8\n\ntake 2 common from the numerator and denominator and cancel them.\n\nx = 2{ -1 +- sqrt(13) } / 4*2\nx = { -1 +- sqrt(13) } / 4\n\nthats your answer.\n\nhave a nice day :)
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
5554
Solving the problem stepwise -\n(x - 8 ) ( x - 2 )\n= x(x-2) - 8(x-2)\n=x^2-2x-8x+2\n=x^2-10x+2\n\nThese two sites will help you for the concepts.
922434
There is only one step. Take the quadratic formula and plug in the numbers.\n\nIf ax^2 + bx + c = 0,\nx = (-b +/- sqrt(b^2-4ac))/2a\n\nFor the equation you gave, a=1, b=19, c=88. Plug those numbers in and see what you get.\n\nIf you're confused by the +/-, it's shorthand for writing two things in one. There are usually two values of x that work, one with + and one with -.
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
236017
sorry i don't know mathematica but you can solve it easily using matlab. Enter rthe coefficient matrix as follows for example,\nA=[1 2 3 4; 2 3 3 4;........]\nand the right hand side\nb=[ 2 3 4 4]\nonly for illustration. ";"indicates columns.\nA/b solves this system and gives the solution vector x.
952497
Ok, I assume you know the quadratic equation, but just for convenience I'm going to write it out:  x = (-b +/- root(b^2 - 4ac))/2a.  \n\n1) x = (5 +/- root(25 + 24))/4.  (Remember, you have to subtract the 3 from each side to get the equation to equal 0, so c is -3.)  That simplifies to x = (5 +/- 7)/4--->x = 12/4 or -2/4--->x = 3 or -1/2.\n\n2) x = (2 +/- root(4 - 12))/6.  Are you sure that wasn't supposed to be a negative 3x^2 or minus 1?  If it were, the term inside the square root would be 16, a positive perfect square.  As it is, it's -8, whose square root is 2root(2)i.  That seems pretty complex (no pun intended) compared to the other one.  \n\nI'm going to solve that one assuming that it should have ended in -1 rather than 1.  Then x = (2 +/- root(4 + 12))/6--->x = (2 +/- 4)/6--->x = -2/6 or 6/6--->x = -1/3 or 1.
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!
848564
Substitute 2x-1 for y in the middle part, you get:\nx^2 + x -6(2x-1) = 2x-1\nThis you can reduce to a quadratic equation for x.
530142
use the quadratic equation:\nx=-b +/- the square root of b^2 - 4ac all over 2a\na=1,b=7,c=10 so plug it in\nx=-7 +/- the square root of 49-40 all over 2\n-7 +3 = -4 divided by 2 x=-2\n-7-3=-10 divided by 2 = -5\nso x = -2 or x=-5