what is the quartic equation given that the quartic roots are (2x +3 )(x - 2)(x - 1)(3x-2)?
144536
6x^4 - 13x^3 - 9x^2 + 28x - 12
886791
(px+2)(3x+q) = rx ^2 + x - 2\n\n(3p)*x^2 + (pq+6)*x + 2q =  rx ^2 + x - 2\n\nComparing coefficients on both side: we have\n\n2q = -2 , q = -1\n\npq+6 = 1, p(-1) + 6 =1 , p = 5\n\n3p = r , r = 3*5 = 15\n\nTherefore the solution is p = 5, q = -1 and r = 15\n\nHope it helps!
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.
898169
b = -4\nc = 1\n\nIn any quadratic ax^2 + bx + c = 0,\nThe sum of the roots = -b/a\nthe product of the roots = c/a\nHere, a = 1, so the sum of the roots = -b\nthe product of the roots = c.\n\nSo right away we know b = -4.\nIf the roots are p and q, then\np + q = 4\np - q = 2√3  { I assume this is the number you were trying to represent and not √(2.3) }\nSo, 2p = 4 + 2√3  \np = 2 + √3\nq = 2 - √3\nand pq = 1, so c=1
233017
the number of roots for any equation is the degree of the equation. here it is 3 (coz the max power or exponent is 3)
976750
f ' (x) = 4x ^ 3 - 18x ^2 + 24 x - 10 = 0\n\nJust by examination, x = 1 is a solution. Knowing that x = 1 is a root, use synthetic division and then factor the remaining quadratic expression to determine that:\n\n4x ^ 3 - 18x ^2 + 24 x - 10 = 2(2x - 5)(x - 1)(x - 1) = 0\n\nThe roots are 1 and 5/2.
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^_^
1185125
√3 (√3+2) is of the form a(b+c)\na(b+c) = (ab+ac)\nso we simplify to \n√3√3+√3*2\n√a√a = √a*a = √a² = a\nso it's 3+2√3.\nThe rules for roots are exactly the same as the rules for exponents, since a (square) root is merely a exponent of (1/2).  A cube root is an exponent of (1/3).\nⁿ√a root is the same as a^(1/n).
877232
If you are looking for an exact answer, be prepared to find the roots of a 4th degree polynomial. a formula can be found at http://planetmath.org/encyclopedia/QuarticFormula.html\n\nx^.5+y=5\nx+y^.5=10\n\nsubstituting z=y^.5 and combining the two equation to cancel out x and the new quartic equation to solve is z^4-10z^2+z+15=0\nplugging in the formula in the link will give you z=1.44024\nso y=2.07429 making x=8.55976\n\nas for approximate answers, there are various methods, bisection, newton's method. you can try graphing them to see where to begin looking.
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
800627
x = 3/(x-2)\n\n=> x(x-2) = 3\n=> x^2 - 2x = 3\n=> x^2 - 2x - 3 = 0\n=> (x-3) (x+1) = 0\n\nEither x-3 = 0 => x=3\nOr x+1 = 0 => x=-1
888899
f(x) = x^2 + 3x\n\nf(2-a) = (2-a)^2 + 3(2-a)\n\nExpand (2-a)^2 to (2-a)(2-a) and distribute the 3(2-a)\n\nf(2-a) = (2-a)(2-a) + 6 - 3a\n\nf(2-a) = 4 -2a -2a + a^2 + 6 - 3a\n\nThe solution is:\n\nf(2-a) = 10 - 7a + a^2
681679
The solution involves complex numbers ('i' notation)\n\nThe factorized equation is:\n\n(b + 3 - i)(b + 3 + i)\n\nwhere 'i' is the imaginary complex number = square root of -1\n\n(TOOK ME WHOLE OF 15 MINS!!! MY MATH IS SLIPPING!!!)\n\neinsteinbrainiac - you came close to the solution
982041
Well dear\n\n-If the first root is " α " and the second roots is " α + 3"  so have  \n▪x ² - (α + β) + α β = 0 \n▪( x - α )(x - β ) = 0 \n-b/a = α+ β  \n&\nc/a = α β\nso we have ; \nc/a = m/4  \n-b/a = - ( -16) / 4 = 4 \nα + α + 3 = 4 , 2α + 3 = 4 \nα = 1/2  ,  β = 3 1/2 \n2 /a = α β  =>  m/4 =  1/2 * 7/2  , m = 7\n\nGood Luck dear
994889
Start with a general cubic equation:\nx^3 +ax^2 +bx+c=0.\nFirst, we want to get rid of the x^2 term, so let \nx=y-(a/3).\nIf you plug this in and expand, the y^2 terms will drop out and you will get an equation of the form\ny^3 +Cy+D=0.\nNow, if you let y=u+v (to get an extra bit of freedom), and expand, you get\nu^3 +v^3 +(3uv+C)(u+v)+D=0.\nChoose v so that 3uv+C=0, i.e. v=-C/(3u). Plug this into\nu^3 +v^3 +D=0\nand multiply through by u^3 to get a quadratic in u^3. Solve\nthis by the quadratic formula and take a cube root to find\nu. There will be three cube roots in the complex plane, so find all of them. For each u, find v=-C/(3u), then y=u+v, and finally x=y-(a/3).\n\nThis is often a rather cumbersome proceedure, even if you do some of it ahead of time. Also, the roots may not 'look' like what you expect. If you apply this to an equation that has a root of, say, 4, you will obtain 4 as a sum of two cube roots of quadratic expressions. To recognize the answer as 4 is a challenge in and of itself.
886499
1). f(x) = (x-3)(x²+3x-5) +2. Multiply out to get your answer.\n2). f(1) =1, so we have to solve\nx³-8x²+17x-10 = 0.\nWe know x = 1 is a root of this equation, so x-1 is a\nfactor of the left-hand side. The other factor, obtained\nby long division or synthetic division is\nx²-7x+10.\nNow set this equal to 0 and solve:\n(x-5)(x-2) = 0,\nso the other 2 solutions are x = 5 and x = 2.
78524
(1-a)+3(a-2)-(2a-3)=1