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By the fundamental theorem of algebra, if the monic polynomial equation x 2 + bx + c = 0 has complex coefficients, it must have two (not necessarily distinct) complex roots. Unfortunately, the discriminant b 2 − 4c is not as useful in this situation, because it may be a complex number. Still, a modified version of the general theorem can be ...
One particular solution is x = 0, y = 0, z = 0. Two other solutions are x = 3, y = 6, z = 1 , and x = 8, y = 9, z = 2 . There is a unique plane in three-dimensional space which passes through the three points with these coordinates , and this plane is the set of all points whose coordinates are solutions of the equation.
The roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
The result x 2 is then multiplied by the second term in the divisor −3 = −3x 2. Determine the partial remainder by subtracting −2x 2 − (−3x 2) = x 2. Mark −2x 2 as used and place the new remainder x 2 above it.
In mathematics, a quadratic function of a single variable is a function of the form [1] = + +,,where is its variable, and , , and are coefficients.The expression + + , especially when treated as an object in itself rather than as a function, is a quadratic polynomial, a polynomial of degree two.
The formula for the difference of two squares can be used for factoring polynomials that contain the square of a first quantity minus the square of a second quantity. For example, the polynomial can be factored as follows:
For the folded general continued fractions of both expressions, the rate convergence μ = (3 − √ 8) 2 = 17 − √ 288 ≈ 0.02943725, hence 1 / μ = (3 + √ 8) 2 = 17 + √ 288 ≈ 33.97056, whose common logarithm is 1.531... ≈ 26 / 17 > 3 / 2 , thus adding at least three digits per two terms. This is because the ...
Figure 2: A paraboloid constrained along two intersecting lines. Figure 3: Contour map of Figure 2. The method of Lagrange multipliers can be extended to solve problems with multiple constraints using a similar argument. Consider a paraboloid subject to two line constraints that intersect at a single point. As the only feasible solution, this ...