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and is the positive root of the equation x 2 − x − n = 0. For n = 1, this root is the golden ratio φ, approximately equal to 1.618. The same procedure also works to obtain, if n > 0, = (+ +), which is the positive root of the equation x 2 + x − n = 0.
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.
However, for any degree there are some polynomial equations that have algebraic solutions; for example, the equation = can be solved as =. The eight other solutions are nonreal complex numbers , which are also algebraic and have the form x = ± r 2 10 , {\displaystyle x=\pm r{\sqrt[{10}]{2}},} where r is a fifth root of unity , which can be ...
For example, let a denote a multiplicative generator of the group of units of F 4, the Galois field of order four (thus a and a + 1 are roots of x 2 + x + 1 over F 4. 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 ...
Denoting the two roots by r 1 and r 2 we distinguish three cases. If the discriminant is zero the fraction converges to the single root of multiplicity two. If the discriminant is not zero, and |r 1 | ≠ |r 2 |, the continued fraction converges to the root of maximum modulus (i.e., to the root with the greater absolute value).
If (1 + z) 1/2 = 1 + a 1 z + a 2 z 2 + ⋯ is the binomial expansion for the square root (valid in |z| < 1), then as a formal power series its square equals 1 + z. Substituting N for z, only finitely many terms will be non-zero and S = √λ (I + a 1 N + a 2 N 2 + ⋯) gives a square root of the Jordan block with eigenvalue √λ.
If n is even, a complex number's nth roots, of which there are an even number, come in additive inverse pairs, so that if a number r 1 is one of the nth roots then r 2 = –r 1 is another. This is because raising the latter's coefficient –1 to the nth power for even n yields 1: that is, (–r 1) n = (–1) n × r 1 n = r 1 n.
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.