A quadratic Equation with roots 7+i and 7-i is.... ?
952054
To find the equation, just substitute the 2 roots to the "Quadramagic formula" (...)\nx² - (p + q)x + pq = 0\nwhere p and q are the roots.\n\nHence, p = 7 + i and q = 7 - i.  We substitute:\nx² - [(7 + i) + (7 - i)]x + (7 + i)(7 - i) = 0\n\nWe add and multiply\nx² - (14)x + (7² - i²) = 0\n\nSince i² = -1, then\nx² - 14x + 49 - (-1) = 0\n\nthen,\nx² - 14x + 49 + 1 = 0\n\nTherefore, the quadratic equation is\nx² - 14x + 50 = 0\n\nTherefore, the correct answer is 1).\n\n^_^
870960
Do you mean that the roots of x^2+pX+r are α  and β and you need to find quadratics in terms of p and q?\nIf so you can use the following:\nIn any quadratic ax^2 + bx + c = 0, the sum of the roots = -b/a and the product of the roots = c/a.\nUsing this,  -(α + β) = p, and αβ= r.\n\nAnswers are for terms b and c in the quadratic x^2 + bx + c\ni)\nb = -(1/α + 1/β) = -(α + β)/αβ = p/r\nc = 1/αβ = 1/r\n\nii) \nb = -(1/α^2 + 1/β^2) =  -(α^2 + β^2)/(αβ)^2 = -[(α + β)^2 - 2αβ])/(αβ)^2 = -(p^2 - 2r)/r^2\nc = 1/(αβ)^2 = 1/r^2\n\niii) \nb = (α^2β + αβ^2) = αβ(α + β) = -pr\nc = -α^2β * -αβ^2 = (αβ)^3 = r^3
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x
1151963
Yes it is\nX+7=11\n  -7    -7\nX=4
830938
Sqrt(2y+7) + 4 = y\n>>\n2y + 7 = (y - 4)^2\n>>\n(y - 4)^2 -2y -7 = 0\n>>\n(y - 4)^2 -2(y - 4) -15 = 0\n\ny - 4 = (2 + sqrt(4 +60))/2 or  (2 - sqrt(4 +60))/2\n>>\ny = 9 or 1\n\nBut if you require the sqrt(2y + 7) be positive, the only acceptable solution is 9.
719816
Round roots??? V(-1) = i
82007
Just type in quadratic equation and maybe roots in a serch engine. You will have your pick of dozens of sites.
605879
(x+3)(x+12)(x+2)(x+8)= [(x+3)(x+8)][(x+12)(x+2)]\n                     = (x^2+11x+24)*(x^2+14x+24)\nlet t=x^2+24+(25X)&#92;2 . \nSo we have: x^2+11x+24=t-(3x)&#92;2 and x^2+14x+24=t+(3x)&#92;2                                 \nso the equation becomes: [t-(3x)&#92;2][t+(3x)&#92;2] = 4x^2\n                    <-->t^2-(9x^2)&#92;4=4x^2  \n                    <-->t^2=(25x^2)&#92;4\n                    <-->t=(5x)&#92;2 or t=-(5x)&#92;2\n                    <-->x^2+10x+24=0 or x^2+15x+24=0\ncontinue by yourself
844359
7x - 30 <19 \n\n7x - 30 + 30 < 19 + 30\n\n7x < 49\n\n7x/7 <  49 / 7\n\nx < 7\n\nThe answer is x < 7\n\nInsert th x value into the equation\n\n- - - - - - - - -\n\nCheck\n\n7x - 30 < 19\n\n7(7) - 30< 19\n\n49 - 30 < 19\n\n19 = 19\n\n- - - - -- - -s-
440770
i think u mean (7-i)/(1+i)\n\nto solve this u multiply the top and bottom by the complex conjugate of the denominator, so in this case by (1-i)\n\nso we get (7-i)/(1+i) * (1-i)/(1-i) = (7-8i+i^2)/(1-i^2)\n\nwe know i^2 = -1\n\nso answer is (6-8i)/2 = 3-4i
673027
In order to find a polynomial equation with real coefficients that has roots 3-i, you need to understand that if 3-i is a root, then, 3+i is also a root because 3+i is the conjugate of 3-i. Then, you should multiply (3-i) with (3+i) using foil, or, distributive properties. You also need to understand that the imaginary number i\n\n\n(3-i)(3+i) 9+3i-3i-i² = 9-(-1)=10
1150325
The quadratic formula is:  [-b+/-((b^2-4ac))/(2a)] where a is the number befor x^2, b is the number before x and c is the integer.    Using this formula we get:\n\n1. x = [1+/- (1-(4)(6)(-12))^1/2/((2)(6))]\n    x = [1+/- 17]/(12)\n    x = 18/12 or -16/12\n    x = 1.5, x = -4/3\n\n2. x =  [6+/- (36-(4)(1)(13))^1/2/((2)(1))]\n    x = [6 +/- (16)^(1/2) i}(2) {since there is a negative sign inside the sq root we replace it by and positive sq root and imaginary number}.\n    x = [6 +/- 4i]/(2)\n   x = 3 + 2i, x = 3-2i
10571
50 is right.\n{i^2 -(-1)^i | i=1,2,...}
968624
1.  \n\n7x + 5i = 7 - 6yi\n\nIsolate x:  \n\n7x = 7 - 6yi - 5i\n\nSolve the right hand side in the proper form of complex numbers (a + bi)\n\n7x = 7 + i(-6y - 5)\nx = 1 + ( (-6y - 5)/7 ) i\n\nYour goal is to get some value for a (the number without the i) and some value for b (the number next to the i).  That's why I factored out the i so that we get the complex number proper form.\n\n2.\n\nThe quadratic formula for ax^2 + bx + c = 0 goes as follows:\n\nx = (-b +/- sqrt(b^2 - 4ac))/2a\n\nIn the above question, a = 2, b = 7, and c = 6.  All you need to do is substitute.  Note that "+/-" means "plus or minus", meaning you'll get two solutions for x.
1113001
take squre root of both sides\nx-1=+ or - sqrt(7)\n\nthen \nthe first root \nx1=1-sqrt(7)\n\nthe second root\nx2=1+sqrt(7)\n\nBy factoring method:\n\ntake 7 to left\n(x-1)^2-(sqrt(7))^2=0\nfactor like a^2-b^2=(a-b)(a+b)\n\n[x-1-sqrt(7)] * [x-1+sqrt(7)]=0\n0*0=0\nx=1+sqrt(7)\nor\nx=1-sqrt(7)
301645
If you have an equation like 5x^2 + 7x - 6 = 0:\n\nGet the x^2 term by itself\nx^2 + 7x/5 - 6/5 = 0\n\nMove the constant to the other side\nx^2 + 7x/5 = 6/5\n\nTake half of the coefficient for the x term and square it: half of 7/5 is 7/10, squared is 49/100. Add this to both sides of the equation\n\nx^2 + 7x/5 + 49/100 = 6/5 + 49/100\nx^2 + 7x/5 + 49/100 = 169/100\n\nThe left side is a perfect square trinomial - factor it\n(x + 7/10)^2 = 169/100\n\nTake the square root of both sides and solve for x\nx + 7/10 = +/- 13/10\nx = -7/10 +/- 13/10\nx = 6/10 or -20/10\nx = 0.6 or -2
932849
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