How do you factor a quadratic equation?\nhow do you solve a quadratic equation by completeing the square?
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Factoring Quadratic Equations\n\n Certain  complete  quadratic  equations  can  be  solved  by  factoring.   If  the  left-hand  side  of  the general form of a quadratic equation can be factored, the only way for the factored equation to be  true  is  for  one  or  both  of  the  factors  to  be  zero.    For  example,  the  left-hand  side  of  the quadratic  equation  x2  +  x  -  6  =  0  can  be  factored  into  (x  +  3)(x  -  2).   The  only  way  for  the equation (x + 3) (x  - 2) = 0 to be true is for either (x + 3) or (x - 2) to be zero.   Thus, the roots of quadratic equations which can be factored can be found by setting each of the factors equal to zero and solving the resulting linear equations.  Thus, the roots of (x + 3)(x - 2) = 0 are found by setting x + 3 and x - 2 equal to zero.   The roots are x = -3 and x = 2. \n\nFactoring  estimates  can  be  made  on  the  basis  that  it  is  the  reverse  of  multiplication.    For example, if we have two expressions (dx + c) and (cx + g) and multiply them, we obtain (using the distribution laws)\n\n (dx + c) (fx + g) = (dx) (fx) + (dx) (g) + (c) (fx) + cg = = dfx2 + (dg + cf)x + cg. \n\nThus, a statement (dx + c) (fx + g) = 0 can be written \n\ndf x2 + (dg + cf)x + cg = 0.\n\n Now, if one is given an equation  ax2  +  bx  +  c  = 0, he knows that the symbol  a  is the product of two numbers (df) and c is also the product of two numbers.   For the example 3x2 - 4x - 4 = 0,  it  is  a  reasonable  guess  that  the  numbers  multiplying  x2  in  the  two  factors  are  3  and  1, although they might be 1.5 and 2.   The last -4 (c in the general equation) is the product of two numbers (eg), perhaps -2 and 2 or -1 and 4.  These combinations are tried to see which gives the proper value of b (dg + ef), from above. \n\nThere are four steps used in solving quadratic equations by factoring. \n\nStep 1. Using  the  addition  and  subtraction  axioms,  arrange  the  equation  in  the general quadratic form ax2 + bx + c = 0. \n\nStep 2. Factor the left-hand side of the equation. \n\nStep 3. Set each factor equal to zero and solve the resulting linear equations. \n\nStep 4. Check the roots.\n\nNow Completing the Square:\n\nx2 + 6x + 9 = 2       Trinomial Square\n(x + 3)(x + 3) = 2   (A + B)2 = A2 + 2AB + B2\n(x + 3)2 = 2         (A - B)2 = A2 - 2AB + B2\n\nie.\nx2 + 14x + 49 = (x + 7)2    (14/2)2 = 72 = 49\n x2 + 6x + 9 =  (x + 3)2    (6/2)2 = 32 = 9\nx2 + 10x + 25 = (x + 5)2    (10/2)2 = 522 = 25\nx2 - 14x + 49 = (x - 7)2    (-14/2)2 = (-7)2 = 49\n x2 - 6x + 9 =  (x - 3)2    (-6/2)22 = (-3)22 = 9\nx2 - 10x + 25 = (x - 5)2    (-10/2)2 = (-5)2 = 25\n\nWhen solving an equation, to complete the square of an expression like x2 + bx, we take half the x coefficient, which is b/2, and square. Then we add that number, (b/2)2, on both sides\n\nExample 9: Solve x2 - 6x + 8 = 0\n\nExample 11: Solve 2x2 = 3x - 7\n\nx = 3/4 ± -47/16
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While looking for easier ways to do things with factors, I realized there might be a commonality between the left side and the right side.\n\nfactor a 2 out of the left side...\n2(17x-3.8) = (17x-3.8)(x+12.5)\n\ndivide both sides by (17x-3.8)...\n2 = x+12.5\n\nsubtract 12.5 from both sides...\n-10.5 = x\n\nAnd there you have it.  x = -10.5\n\n\nAnother possible approach would have been to do what the second answerer suggested, but use "completing the square" instead of the quadratic formula.
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The quadratic formula goes: (-b ± √b² - 4ac)/ 2a\nSo x²=a, -6x=b, -7=c\nNow plug it in:\n [-(-6) ± √(6)² -4(1)(-7)]/ 2(1) \nsimplify: (6± √36 + 28)/ 2\nsimplify further: (6±√64)/2\nwhich equals: (6±8)/2\nand if you separate it out: (6-8)/2 = -1      and   (6+8)/2 = 7
213632
Here's how to complete the square:\n\nI'll use the example of \n\n2x^2 - 4x -6 = 0\n\n1.  Divide all the terms by the coefficient of the x^2 term.\n\nThe coefficient of the x^2 term in your example is 2, so\n\n(2/2)x^2 - (4/2)x -(6/2) = 0\nx^2 - 2x -3 = 0\n\n2.  Bring the constant term to the right side of the equation.\n\nx^2 - 2x = 3\n\n(This is where you are at in your example)\n\n3.  Complete the square\n\nAny square of the form (x + a)^2 is x^2 + 2a + a^2.  Look at the x term in your example.  The coefficient is -2.  In your example, 2a is going to be -2.  So a=-1.  Right now, you have (on the right side)\n\nx^2 +2(-1)\n\nThe missing part of the square is (-1)^2, which is equal to 1.  You complete the square by adding that value to the right side.  You keep the euqation balanced by adding it to the left side:\n\nx^2 - 2x +1 = 3+1\n(x - 1)^2 = 4\n\n4.  Take the square root.  Remember that there are two possibilities:\n\nx - 1 = 2\nx - 1 = -2\n\n5.  Solve both equations:\n\nx = 3\nx = -1\n\n6.  Check both your answers.  Anybody in the world could make an error here.  The most important thing is to check both of your answers:\n\nx = 3\n\n2x^2 - 4x -6 = 0?\n2(9) -4(3) -6 = 18 -12 -6 =0\n\nx = -1\n\n2x^2 - 4x -6 = 0?\n2(1) -4(-1) -6 = 2 + 4 - 6 = 0
519998
Is this one problem?  If so, you can solve it like this:\n\n(a/b) / (c/d) = (a/b) * (d/c)\n\n~ ~ ~ ~ \n\n( (x² - 4) / x² ) / ( (x²+5x+6) / x ) = \n\n( (x² - 4) / x² ) * ( x / (x²+5x+6) ) = \n\n( (x² - 4) * x) ) / (x² * (x²+5x+6) ) = \n\n( (x² - 4)  / (x * (x²+5x+6) ) = \n\n{{ now notice you can factor your quadratic equations, to get the following }}\n\n(x+2)(x-2)  / (x * (x+2)(x+3) ) = \n\n(x-2) / (x(x+3))\n\nHope that helped!\n\n~ ♥ ~
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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
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The factorization of the equation you gave is (x + 4)^2 = 0.\n\nOnly 0 squared equals zero, so x + 4 has to equal 0.  By subtracting 4 from each side you get x = -4.\n\nHope this helps!
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20x² - 8x + 4 = 0  Divide both sides by 4.\n5x² - 2x + 1 = 0\n\nTo see if it's factorable, look at the "discriminant" of the quadratic equation:\n\nb² - 4ac = (-2)² - 4(5)(1) = 4 - 20 = -16\n\nSince the discriminant is negative (that's the part under the square root in the quadratic equation), you can say:\n\n-- There is no solution (over the real numbers).\n-- There are two complex conjugate solutions (over the complex numbers).\n\nIf you don't know what complex numbers are, then go with the first answer.
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ok re arranged the second question so that the 6 is factored out so its x= such and such then sub the x equals such and such into the first equation and solve then do the same thing to Y in the second equation
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you need to add +1\nso that the equation can be represented as a perfect square\nx^2 - 2x + 1 = (x - 1) (x - 1) = (x - 1)^2\n\nthis is called solving by perfecting the square\nto do this, make sure that the coefficient of the first term is 1.\nThen, divide the coefficient of the middle term by 2,\nthen square the whole thing.\n\nIn this case, the coefficient of the middle term is -2\n-2 / 2 = -1\n(-1)^2 = +1
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x= 3/5\n\nIf you factor by grouping you would get\n25x^2 - 15x -15x + 9 = 0\n\nthen you would factor the first two terms and the last two terms\n5x(5x-3)   -3(5x-3)    =  0\n\nuse the terms outside the parentheses are one factor and the one inside the parentheses as the second factor\n(5x-3)(5x-3) = 0\n\nset to zero\n5x-3 = 0\n\ndo algebra\nx=3/5\n\nor you could have just noticed that that equation is a perfect square
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x^4 + 4\n\nWell, you can add 0 in the form of +a - a, so\nx^4 + 4 +4x^2 - 4x^2\n\nRegrouping them, we get\nx^4 + 4x^2 + 4 - 4x^2\n\nUsing factoring, we get\n(x^2 + 2)^2 - (2x)^2\n\nThen,\n[x^2 + 2 + 2x][x^2 + 2 - 2x]\n\nRearranging them for a more beautiful sight, we get\n{x^2 + 2x + 2)(x^2 - 2x + 2)\n\nThanks For the Problem ^_^
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I enjoy algebra, and this is easily my favorite type of problem.  So I hope you, too, one day enjoy them for their inner beauty.\n\nFor now, let's tackle this "factoring problem."  We may wish to use the method of "synthetic division" which is arguably easier than "long division of polynomials."\n\nThink back to grade school where you learned that 6 = 2 * 3.  This simple statement enabled you to 'deconstruct' the number 6 as either "two times three" or "three times two".  Even if you forgot one of the 'factors' you could conjure it up again quickly by asking yourself, "What is six divided by two?" for example.\n\nIn problem (a), you are essentially asking, "What are the coefficients of P(x) if the only factor we know is (x-1)?"\n\nLet's set up synthetic division (see link below), by placing all the coefficients of P(x), and using 1 (the zero implied by (x-1)) as the divisor:\n\n1 ||  a  b  c  d\n___________\n.....a\n\n1 ||  a  b  c  d\n_____a_____\n.....a (a+b)\n\n1 ||  a  b        c       d\n_____a___(a+b)____\n.....a (a+b) (a+b+c)\n\n1 ||  a  b        c                d\n_____a___(a+b)___(a+b+c)_\n.....a (a+b) (a+b+c) (a+b+c+d)\n\nNote that this last term must equal 0 since it is assumed that (x-1) is a factor ... thus, the "remainder" term must equal zero!\n\nThere's the result you were after:  a+b+c+d = 0\n\nHope this helps you tackle the others!
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complete the square:\n\ny= -2x^2-4x-7\n\ny= -2(x^2 + 2x + 7/2)\ny = -2(x^2 + 2x + 7/2) + -2(-5/2) + 2(-5/2)\ny = -2(x^2 + 2x + 7/2 -5/2) + 2(-5/2)\ny = -2(x^2 + 2x + 1) - 5\ny = -2(x + 1)^2 -5\n\nThe other way to do it is to use the formula:\nax^2 + bx + c = a(x + (b/2a))^2 + c - (b^2/4a)\n\nwhich also yields -2(x + 1)^2 - 5.\n\nYour answer is correct.
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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.
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-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
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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.