Solving Quadratic Equations?
813596
Okay, the goal here is to get values of x that fit the equation.  These are not like easy linear ones, they have x^2 in them.  We can still solve them.  \n\nx^2=16\nWhat is the reverse sign of the square?  It's the square root.  Square root both sides and get\nx=sq(16)\nSo x=+/-4\nBoth work because (-4)^2 becomes 16 as well.  Those are your answers.\n\n\n2x(x+3)=0\nThe goal here is to make this equation equal zero.  Do you see how there are two "factors?"  Things that are multiplied together?  Well, to equal zero, one of them must equal zero.  Either one can equal zero, so you have different solutions that work.\n\nTake 2x:\n2x=0\nx=0\n\nNow x+3:\nx+3=0\nx=-3\nSo\nx=0,-3\n\n\nHere we have to "factor" this equation:  Find two things that, when multiplied together, equal this.  y^2 and 8y both have y, so factor out the y and get:\ny(y+8) \nWe can do as we did before, and get\ny=0\nand \ny+8=0, so\ny=-8\nOur answers are\ny=0,-8\n\nThe next two are harder:\n\nFirst bring the 28 over to the other side:\n\nx^2+3x-28=0  \n\nThe factored form of this actually has two factors, but none with just x.  To factor this, we need two factors of -28.  When one of them is multiplied by x and added to the other one multiplied by x, they equal 3x.  Now, a negative times a positive is a negative so we are looking for one positive and one negative factor of 28.  Let's try these two:\n\n7 and -4.\n7(-4)=-28\n7x+(-4x) = 7x-4x=3x!\n\n-4 and 7 are our factors, so:\n(x+7)(x-4) is our factored form.  Solve the components:\nx+7=0\nx=-7\n\nx-4=0\nx=4\n\n x=-7,4 are our two answers.\n\n\n\n2x^2+9x-5\n\n\nThe next one is similar, but more complicated.  We have a 2x^2.  No matter.  We need to find two factors of -5, and one of them will be multiplied by 2x instead of x.  The only factors of -5 are +/-1, and +/-5.  It looks like the 5 will be positive since 9x is fairly big and positive.  Let's try -1 and 5:\n2x(5)=10x\nx(-1)=-x\n10x+(-x)= 10x-x=9x!\n\nSo 2x will be multiplied by 5, and x by -1.  In factored form, this is:\n\n (2x-1)(x+5)\n\nBecause you use FOIL (first outside, inside last) so the 2x is not multiplied by -1, but by 5.  Now solve these components.\n\n2x-1=0\n2x=1\nx=1/2\n\nx+5=0\nx=-5\nSo\n x=1/2, -5 are our two answers. \n\nAll done.  Hope this helps.
617358
x = (-b +- sqrt(b^2-4ac))/(2a)\n\nThis equation solves the quadratic equation, where\n\nax^2 + bx + c = 0.
1172065
if it is a quadratic (ax^2 + bx + c) then put it into the quadratic equation. that automatically will give you 0, 1, or 2 real answers. the equation is: x = [-b +/- (square root of b^2 - 4*a*c)] / 2a\n\nif (b^2 - 4*a*c) is negative, you have no real solutions. if it is zero, you have one real solution. if it is positive, you have 2 real solutions.\n\nthen, if for example you get  x=4 and x= -2, your factors are (x-4) and (x+2)
475755
let cos (2x) = y. this forms a normal quadratic equation that you can solve: \n4y^2-4y +1 = 0\n(2y-1)^2=0\n2y-1=0\ny=1/2\ncos 2x = 1/2\n2x = +/- 60 + k360\nx = +/- 30 + k180\nnow sub in values of k (k=0,1,2 etc) depending on the domain given for your solutions\nx = -30, 30, 150, 210 etc
146612
A quadratic equation is a second-order polynomial equation in a single variable  \n\n (1) \n\nwith . Because it is a second-order polynomial equation, the fundamental theorem of algebra guarantees that it has two solutions. These solutions may be both real, or both complex. \n\nThe roots  can be found by completing the square, \n\n (2) \n\n (3) \n\n (4) \n\nSolving for  then gives \n\n (5) \n\nThis equation is known as the quadratic formula. \n\nThe first known solution of a quadratic equation is the one given in the Berlin papyrus from the Middle Kingdom (ca. 2160-1700 BC) in Egypt. This problem reduces to solving \n\n   (6) \n   (7) \n\n(Smith 1953, p. 443). The Greeks were able to solve the quadratic equation by geometric methods, and Euclid's  (ca. 325-270 BC) Data contains three problems involving quadratics. In his work Arithmetica, the Greek mathematician Diophantus  (ca. 210-290) solved the quadratic equation, but giving only one root, even when both roots were positive (Smith 1951, p. 134). \n\nA number of Indian mathematicians gave rules equivalent to the quadratic formula. It is possible that certain altar constructions dating from ca. 500 BC represent solutions of the equation, but even should this be the case, there is no record of the method of solution (Smith 1953, p. 444). The Hindu mathematician Aryabhata (475 or 476-550) gave a rule for the sum of a geometric series that shows knowledge of the quadratic equations with both solutions (Smith 1951, p. 159; Smith 1953, p. 444), while Brahmagupta  (ca. 628) appears to have considered only one of them (Smith 1951, p. 159; Smith 1953, pp. 444-445). Similarly, Mahavira (ca. 850) had substantially the modern rule for the positive root of a quadratic. Sridhara (ca. 1025) gave the positive root of the quadratic formula, as stated by Bhaskara (ca. 1150; Smith 1953, pp. 445-446). The Persian mathematicians al-Khwarizmi  (ca. 825) and Omar Khayyám  (ca. 1100) also gave rules for finding the positive root. \n\nViète  was among the first to replace geometric methods of solution with analytic ones, although he apparently did not grasp the idea of a general quadratic equation (Smith 1953, pp. 449-450). \n\nAn alternate form of the quadratic equation is given by dividing (◇) through by : \n\n (8) \n\n (9) \n\n (10) \n\nTherefore, \n\n (11) \n\n (12) \n\n (13) \n\nThis form is helpful if , where  denotes much greater, in which case the usual form of the quadratic formula can give inaccurate numerical results for one of the roots. This can be avoided by defining \n\n (14) \n\nso that  and the term under the square root sign always have the same sign. Now, if , then \n\n (15) \n\n   (16) \n   (17) \n\nso \n\n   (18) \n   (19) \n\nSimilarly, if , then \n\n (20) \n\n   (21) \n   (22) \n\nso \n\n   (23) \n   (24) \n\nTherefore, the roots are always given by  and . \n\nNow consider the equation expressed in the form \n\n (25) \n\nwith solutions  and . These solutions satisfy Vieta's formulas \n\n   (26) \n   (27) \n\nThe properties of the symmetric polynomials appearing in Vieta's formulas then give \n\n   (28) \n   (29) \n   (30) \n\n \nGiven a quadratic integer polynomial , consider the number such polynomials that are factorable over the integers for  and  taken from some set of integers . For example, for , there are four such polynomials, \n\n   (31) \n   (32) \n   (33) \n   (34) \n\nThe following table summarizes the counts of such factorable polynomials for simple  and small . Plots of the fractions of factorable polynomials for  (red),  (blue), and  (green) are also illustrated above. Amazingly, the sequence for  has the recurrence equation \n\n (35) \n\nwhere  is the number of divisors of  and  is the characteristic function of the square numbers. \n\n Sloane factorable over  for , 1, ... \n A067274 1, 4, 10, 16, 25, 31, 41, 47, 57, ... \n A091626 1, 2, 4, 6, 9, 11, 14, 16, 19, 22, ... \n A091627 0, 0, 1, 2, 4, 5, 7, 8, 10, 12, 14, ...
482048
-b (+/-) sqrt(b^2 - 4ac)\n--------------------------------- = x\n2a\n\nUsing the equation "ax^2 + bx + c = 0" to find the values of a, b, and c.
368125
if this is the problem:\n((x+3)/2)*((x^2-x)/(x+3))=21\nthen you solve by\ncancelling out (x+3)\n(1/2)*(x^2-x)=21\nmultiplying by 2\nx^2-x=42\nmaking it quadratic\nx^2-x-42=0\nand using x=(-b(+-)srt(b^2-4ac))/2a\nx=(1(+-)srt(1-4*1*(-42)))/2\nx=(1(+-)srt(169))/2\nx=(1+13)/2  or x=(1-13)/2\nx=7   or x= -6
1100359
1)If it's an equation then solve it quadratically\n\n(X-3)= 0 or (X-3)= 0\n\ntherefore X= 3 (as the 3 moves to the other side making it positive)\n\n2)If you have to find the products ude the distributive law:\n\nXsquared -3X-3X+9\ntherefore the answer would be Xsquared -6X+9\n\nBut I'm sure you needed the Equation one so use my first answer and method.
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!
754366
You need to use the quadratic equation as it is not factorable into rational expressions.\nThe form of the equation is:\nax^2 + bx + c = 0 \nSo in your case,\na = -2, b = 6, c = -3\nand the solution is:\nx = [ -b ± √(b^2 - 4ac) ] / (2a)\n\nso substitute and reduce.\n\nx = [ -6 ± √(6^2 - 4(-2)(-3)) ] / (2(-2))\n\nx = [ -6 ± √(36 - 24) ] / (-4)\n\n= [ -6 ± √12 ] / (-4)\n= [ -6 ± 2√3 ] / (-4) \n= (3 ± √3)/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 :)
589722
Step 1: Solve one equation for one variable in terms of the other.\nStep 2: Plug that into the other equation and solve for the second variable.\nStep 3: Plug the second variable into the first equation and solve for the first variable.
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
219130
I could not get your question. We all know that the quadratic equations are those equation which contain a square with the unknown (x). Quadratic equation will always have two roots. But, where did you get the "Four"? Would you explain a bit more.
730173
x = (-b +/- sqrt(b^2 - 4ac))/2a\n\nwhere a= 1, b= -6 , c=8\n\nx1 = (6 + sqrt(36 - 32))/2 = 4\nx2 = (6 - sqrt4)/2 = 2
1049820
The quadratic formula:\nx= -b(+/-) the square root of (b squared - 4ac), all of this over 2a
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