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To complete the square, form a squared binomial on the left-hand side of a quadratic equation, from which the solution can be found by taking the square root of both sides. The standard way to derive the quadratic formula is to apply the method of completing the square to the generic quadratic equation a x 2 + b x + c = 0 {\displaystyle ...
Given a quadratic polynomial of the form + + it is possible to factor out the coefficient a, and then complete the square for the resulting monic polynomial. Example: + + = [+ +] = [(+) +] = (+) + = (+) + This process of factoring out the coefficient a can further be simplified by only factorising it out of the first 2 terms.
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 .
The quadratic equation on a number can be solved using the well-known quadratic formula, which can be derived by completing the square. That formula always gives the roots of the quadratic equation, but the solutions are expressed in a form that often involves a quadratic irrational number, which is an algebraic fraction that can be evaluated ...
He presented a method of completing the square to solve quadratic equations, sometimes called Śrīdhara's method or the Hindu method. Because the quadratic formula can be derived by completing the square for a generic quadratic equation with symbolic coefficients, it is called Śrīdharācārya's formula in some places.
Solving an equation symbolically means that expressions can be used for representing the solutions. 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.
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