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A solution in radicals or algebraic solution is an expression of a solution of a polynomial equation that is algebraic, that is, relies only on addition, subtraction, multiplication, division, raising to integer powers, and extraction of n th roots (square roots, cube roots, etc.). A well-known example is the quadratic formula
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.
Yet the above logic is still valid to show that if abc = 0 then a = 0 or b = 0 or c = 0 if, instead of letting a = a and b = bc, one substitutes a for a and b for bc (and with bc = 0, substituting b for a and c for b). This shows that substituting for the terms in a statement isn't always the same as letting the terms from the statement equal ...
The solutions of the quadratic equation ax 2 + bx + c = 0 correspond to the roots of the function f(x) = ax 2 + bx + c, since they are the values of x for which f(x) = 0. If a , b , and c are real numbers and the domain of f is the set of real numbers, then the roots of f are exactly the x - coordinates of the points where the graph touches the ...
Geometrically, this says that the solution set for Ax = b is a translation of the solution set for Ax = 0. Specifically, the flat for the first system can be obtained by translating the linear subspace for the homogeneous system by the vector p .
The solution = is in fact a valid solution to the original equation; but the other solution, =, has disappeared. The problem is that we divided both sides by x {\displaystyle x} , which involves the indeterminate operation of dividing by zero when x = 0. {\displaystyle x=0.}
Simplifying this further gives us the solution x = −3. It is easily checked that none of the zeros of x (x + 1)(x + 2) – namely x = 0, x = −1, and x = −2 – is a solution of the final equation, so no spurious solutions were introduced.