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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.
For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1. If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0.
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.
The Basel problem is a problem in mathematical analysis with relevance to number theory, concerning an infinite sum of inverse squares.It was first posed by Pietro Mengoli in 1650 and solved by Leonhard Euler in 1734, [1] and read on 5 December 1735 in The Saint Petersburg Academy of Sciences. [2]
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 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]
A necessary (but not sufficient) condition for solvability is that n is not divisible by 4 or by a prime of form 4k + 3. [note 3] Thus, for example, x 2 − 3y 2 = −1 is never solvable, but x 2 − 5y 2 = −1 may be. [27] The first few numbers n for which x 2 − ny 2 = −1 is solvable are
On the other hand, if such a tiling uses exactly k of the 2 × 1 tiles, then it uses n − 2k of the 1 × 1 tiles, and so uses n − k tiles total. There are ( n − k k ) {\displaystyle {\tbinom {n-k}{k}}} ways to order these tiles, and so summing this coefficient over all possible values of k gives the identity.