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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.
In either case the full quartic can then be divided by the factor (x − 1) or (x + 1) respectively yielding a new cubic polynomial, which can be solved to find the quartic's other roots. If a 1 = a 0 k , {\displaystyle \ a_{1}=a_{0}k\ ,} a 2 = 0 {\displaystyle \ a_{2}=0\ } and a 4 = a 3 k , {\displaystyle \ a_{4}=a_{3}k\ ,} then x = − k ...
defines only one solution (), the so-called singular solution, whose graph is the envelope of the graphs of the general solutions. The singular solution is usually represented using parametric notation, as ((), ()), where = /.
The four roots of the depressed quartic x 4 + px 2 + qx + r = 0 may also be expressed as the x coordinates of the intersections of the two quadratic equations y 2 + py + qx + r = 0 and y − x 2 = 0 i.e., using the substitution y = x 2 that two quadratics intersect in four points is an instance of Bézout's theorem.
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.
Here z is the free variable, while x and y are dependent on z. Any point in the solution set can be obtained by first choosing a value for z, and then computing the corresponding values for x and y. Each free variable gives the solution space one degree of freedom, the number of which is equal to the dimension of the solution set.
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To do so, the different variables in the equation are understood as coordinates and the values that solve the equation are interpreted as points of a graph. For example, if x {\displaystyle x} is set to zero in the equation y = 0.5 x − 1 {\displaystyle y=0.5x-1} , then y {\displaystyle y} must be −1 for the equation to be true.