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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.
In mathematics, a quadratic equation (from Latin quadratus 'square') is an equation that can be rearranged in standard form as [1] + + =, where the variable x represents an unknown number, and a, b, and c represent known numbers, where a ≠ 0. (If a = 0 and b ≠ 0 then the equation is linear, not quadratic.)
To convert the standard form to factored form, one needs only the quadratic formula to determine the two roots r 1 and r 2. To convert the standard form to vertex form, one needs a process called completing the square. To convert the factored form (or vertex form) to standard form, one needs to multiply, expand and/or distribute the factors.
This equation states that , representing the square of the length of the side that is the hypotenuse, the side opposite the right angle, is equal to the sum (addition) of the squares of the other two sides whose lengths are represented by a and b. An equation is the claim that two expressions have the same value and are equal.
In elementary algebra, completing the square is a technique for converting a quadratic polynomial of the form + + to the form + for some values of and . [1] In terms of a new quantity x − h {\displaystyle x-h} , this expression is a quadratic polynomial with no linear term.
The quadratic formula =. is a closed form of the solutions to the general quadratic equation + + =. More generally, in the context of polynomial equations, a closed form of a solution is a solution in radicals; that is, a closed-form expression for which the allowed functions are only n th-roots and field operations (+,,, /).
Using Cartesian coordinates in three dimensions, let x = (x, y, z) T, and let A be a symmetric 3-by-3 matrix. Then the geometric nature of the solution set of the equation x T Ax + b T x = 1 depends on the eigenvalues of the matrix A. If all eigenvalues of A are non-zero, then the solution set is an ellipsoid or a hyperboloid.
For example, antiderivatives of x 2 + 1 have the form 1 / 3 x 3 + x + c. For polynomials whose coefficients come from more abstract settings (for example, if the coefficients are integers modulo some prime number p , or elements of an arbitrary ring), the formula for the derivative can still be interpreted formally, with the coefficient ...