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Because (a + 1) 2 = a, a + 1 is the unique solution of the quadratic equation x 2 + a = 0. On the other hand, the polynomial x 2 + ax + 1 is irreducible over F 4, but it splits over F 16, where it has the two roots ab and ab + a, where b is a root of x 2 + x + a in F 16. This is a special case of Artin–Schreier theory.
For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2(y + 1) – 1, a true statement. It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1.
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 imaginary unit or unit imaginary number (i) is a mathematical constant that is a solution to the quadratic equation x 2 + 1 = 0. Although there is no real number with this property, i can be used to extend the real numbers to what are called complex numbers, using addition and multiplication. A simple example of the use of i in a complex ...
To solve this kind of equation, the technique is add, subtract, multiply, or divide both sides of the equation by the same number in order to isolate the variable on one side of the equation. Once the variable is isolated, the other side of the equation is the value of the variable. [37] This problem and its solution are as follows: Solving for x
The solutions –1 and 2 of the polynomial equation x 2 – x + 2 = 0 are the points where the graph of the quadratic function y = x 2 – x + 2 cuts the x-axis. In general, an algebraic equation or polynomial equation is an equation of the form =, or = [a]
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 ...
If an equation P(x) = 0 of degree n has a rational root α, the associated polynomial can be factored to give the form P(X) = (X – α)Q(X) (by dividing P(X) by X – α or by writing P(X) – P(α) as a linear combination of terms of the form X k – α k, and factoring out X – α. Solving P(x) = 0 thus reduces to solving the degree n – 1 ...