I got a pretty bad grade on my test in ALG. 1 and I need a thorough explanation on completing the squares. \n\nHere is one of the questions I missed:  x^2 - 2x = 3
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
531012
Hi again \n\nY = √2x – 3 \n2x – 3 = y ²\nX =( y ² + 3 )/ 2\nf –¹ (x)  =( y ² + 3 )/ 2  =>   f –¹ (x)  = (x ² + 3 )/ 2   &  x ≥ 0 \n\nGood Luck \n try to solve them yourslef again if you got wrong , am sure you can solve them correctly , if you just try your best
349671
You have to complete the square.\n\ny = a(x^2 + b/a x) + c\ny = a(x^2 +b/a x + b^2/4a^2) + c - b^2/4a\ny = a(x + b/2a)^2 + (4ac - b^2)/4a\n\n[4ay - (4ac - b^2)]/4a = a(x + b/2a)^2\n[4ay - 4ac + b^2]/(4a^2) = (x + b/2a)^2\n\n+/- sqrt(4ay - 4ac + b^2)/2a = x + b/2a\n\nx = (-b +/- sqrt(4ay - 4ac + b^2))/2a
105050
Have patience :)  You got some answers the first time you asked this.  Here it is again:\n\nIt does not get eliminated, it gets split up into the 2 factors.\n\nOn this one, you can go ahead and divide by 2 to make things a little easier. You'll be left with\n\n3x^2 - 7x - 6 = 0\n\nNow to factor. You need 2 numbers whose product equals 3*-6, or -18 and whose sum equals -7. From this point on it's just sort of educated guessing. \n\n1*-18 = -18, but 1-18 = -17, so that won't work\n2*-9 = -18, and 1-9 = -7 -- THIS ONE WORKS!! yay!\n\nSo now you can rewrite your problem as\n\n3x^2 - 9x + 2x -6 = 0 and group the first 2 and last 2 terms\n\n(3x^2-9x) + (2x-6) = 0 -- you can factor a 3x out of the first group, and a 2 out of the second\n\n3x(x-3) + 2(x-3)=0 -- and now you can factor an (x-3) out\n\n(x-3)(3x+2)=0\n\n~~~~~~~~~~~~~~~~~~~\nOops, you're right. I was typing too fast.  Oh well, I still got the right final answer :)  Thanks
380222
x^2-x-6>0\nor, x^2-3x+2x-6>0\nor, x(x-3)+2(x-3)>0\nor, (x-3)(x+2)>0\nor x>3 and x<2\n\nhope this helps :)
452660
This answer refers to the problem of y>x^2-2x+3 or verbally\n    y is greater than x squared minus 2 times x plus three\n\nThis problem can be solved through direct substitution.\nYou need to remember that your ordered pairs are in the form of (x,y)\n\nLet's substitute in (1,4).  We know that x=1 and y=4:\n\n4 > 1^2 - 2(1) +3\n4 > (1)(1) -2(1) +3\n4 > 1 - 2 + 3\n4 > -1 +3\n4 > 2            \n\nsince this is true, the correct answer should be:  c  (1,4)\n\nHope this helps!
14593
(x-2)/(x^2-2x-15)+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(-2-x)(x+3)/(x-5)(x+3)\n\n[(x-2) + (x+3)(-2-x)] / [(x+3)(x-5)]\n\nANSWER :\n(x^2+4x+8)/((x+3)(5-x))
535276
if by this you mean\n\nx^2 - 2x - 2 = 0\n\nx^2 - 2x = 2\nx^2 - 2x + 1 = 3\n(x - 1)^2 = 3\nx - 1 = sqrt(3)\nx = 1 ± sqrt(3)\n\n-------------------------\n\nbut if you mean\n\n7x^2 - 2x - 2 = 0\n7x^2 - 2x = 2\nx^2 - (2/7)x = (2/7)\nx^2 - (2/7)x + (1/49) = (63/49)\n(x - (1/7))^2 = (9/7)\nx - (1/7) = (3/7)sqrt(7)\nx = (1/7) ± (3/7)sqrt(7)\nx = (1/7)(1 ± 3sqrt(7))
522445
sq=squared\n(2x+1)(x+2)=5x+2\n\n2xsq+4x+x+2=5x+2  timesd left side\n2xsq+5x+2=5x+2      \n2xsq+5x+2-5x=2     put 5x to other side\n2xsq+2=2               divided by2\nxsq+1=1\nya xsq = 0?
277691
First, FUNNYPLAYA12'S answer is a partial of what I got, and my formula looked like this:\n\n-8X^3 + 10X^2 - 2X = 0\n-----------------------------------\nFactor out one X and you get:\n\n-8X^2 + 10X - 2 = 0\n\nFactor this and FOIL-check\n\n(-4X + 1)\n(2X - 2)\n___________\n-8X^2 (8X + 2X) -2\n\n-8X^2 +10X - 2\n\nX = {1/4, 1)\n\n\nTo verify:\n\n-8(1) + 10(1) -2(1) = 0\n\n-8 + 10 -2 = 0; and finally:\n\n10 - 10 = 0, which is very true!\n----------------------------------------\n\n-8(1/4) + 10(1/4) -2(1/4) = 0\n\n-2 + 2.5 - 0.5 = 0\n\n-2.5 + 2.5 = 0, which is also very true!\n\n\nOTBPOC
440861
Completing the square involves adding constants to both sides of your equation so you have a form:\n\n(x - a)^2 = b\n\nOnce you get this form, you can take the square root of both sides:\n\nx - a = sqrt(b)\n\nx = a + sqrt(b) \n\nSince a square root can be either positive or negative, it is just as correct to write:\n\nx = a - sqrt(b)\n\nNow, back to the original "perfect square"\n\n(x - a)^2 = b\n\nExpanding this out:\n\nx^2 - 2*a + a^2 = b\n\nMoving b to the other side:\n\nx^2 - 2*a + a^2 - b = 0\n\nYour equation is 2*x^2 - x - 2 = 0\n\nForst thing is to get plain x^2, so divide by 2:\n\nx^2 - (1/2) * x - 1 = 0\n\nNow from the form above, a = 1/.4 because 2*a = (1/2)*x\nNext, look at the constant term: \n \n-1 = a^2 - b\n-1 = (1/4)^2 - b\n\nSolving for b: b = 17/16\n\nSo your perfect square equation is:\n\n(x - 1/4)^2 = 17/16\n\nSolving for x:\n\nx - 1/4 = sqrt(17/16)\n\nx = 1/4 + sqrt(17/16)\nx = 1/4 - sqrt(17/16)\n\nSimplifying slightly:\n\nx = (1 + sqrt(17))/4\nx = (1 - sqrt(17))/4
329228
start: (a^2+2a+1)(a+1)\n\nyou basically multiply all the terms\n\na^2(a)+a^2(1)+2a(a)+2a(1)+1(a)+1(1)\n\nwhich is a^3+a^2+2a^2+2a+a+1\n\nwhich is a^3+3a^2+3a+1\n\ntoo bad i've completely forgotten how to go the other way...
483354
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!
330761
i'd love to help but i'm having trouble with it myself\n\nuse ur math textbook it will have it unless ur math book say algebra and is red ( that my alg. book and it doesn't help) \n\nSRY good luck
502457
I would square the x+4 first. \n2(x^2+8x+16)-3(2x-3)(x+7)\n\nthen, distribute the 2:\n2x^2+16x+32-3(2x-3)(x+7)\n\nthen multiply (2x-3) by (x+7) *FOIL\n2x^2+16x+32-3(2x^2+14x-3x-21)\n\nthen combine the x's\n2x^2+16x+32-3(2x^2+11x-21)\n\nthen distribute the 3 *make sure you change the signs since it's -3:\n2x^2+16x+32-6x^2-33x+63\n\ncombine like terms:\n-4x^2-17x+95\nand there you go
538191
4x^2 is the key.\n\nYou get 1 x^2 by multiplying 1 by x^2, so kx times x must equal 4-1=3.\n\nIn the same way for the second:\n\nYou get -1 x^2 by multiplying -1 by x, so 2*k*x^2 = -5-(-1)x^2\n\n2*k*x^2 = -4x^2\nk=-4/2\nk=-2.\n\nThe second is similar.  Basically, look at the x^2 coefficient and look at how many you have through the terms NOT including k, then subtract the number of other ones from the total number of x^2.  kx must be multiplied by x to get x^2, so take into account the other x coefficient.\n\ni.e. (2x)(kx) = 2kx^2
426622
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.