Completing a Square and Quadratic Formula
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The importance of the investigations connected with the expression ax² + bx + c, can hardly be overrated, at least to those students who pursue mathematics to any extent. In the higher branches, great familiarity with these results is indispensible. |
| A. De Morgan (1806-1871)
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By the Fundamental Theorem of Algebra, every polynomial Pn(x) of degree n with complex coefficients has n (perhaps repeated) complex roots. Here we are concerned with the case of n = 2 and the polynomials of second degree. Such polynomials are in the form
where a is assumed not to be zero: a ≠ 0. A quadratic polynomial is assured to have two roots, say, x1 and x2, such that it admits a decomposition
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P(x) = ax² + bx + c = a(x - x1)(x - x2).
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The two roots can be found by means of what is known as a Quadratic Formula:
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x1,2 = (-b ± √b² - 4ac) / 2a.
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The formula is derived by the process of completing a square based on a well known (and easily verifiable) algebraic identity
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(u + v)² = u² + 2uv + v².
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So, for example, if we can identify an expression as u² + 2uv, then to complete the square we need to add to it v² which converts the sum of two terms u² + 2uv into a single square (u + v)².
We start with the quadratic equation ax² + bx + c = 0 and move the free term to the right:
The left hand side can be seen to be an incomplete square in the form u² + 2uv. Indeed,
| | ax² + bx | = (√a x)² + 2(√a x)(b / 2√a) |
| | | = (√a x)² + 2(√a x)(b / 2√a) + (b / 2√a)² - (b / 2√a)². |
(Here u = √a x and v = b / 2√a). We can continue to transform the quadratic equation step by step:
ax² + bx + c = 0,
ax² + bx = -c,
(√a x)² + 2(√a x)(b / 2√a) + (b / 2√a)² = - c + (b / 2√a)²,
(√a x + b / 2√a)² = - c + (b / 2√a)²,
(2ax + b)² = - 4ac + b²,
2ax + b = ± √b² - 4ac,
2ax = ± √b² - 4ac - b,
x = (-b ± √b² - 4ac) / 2a.
Example: solve 2x² - 7x + 3 = 0, with a = 2, b = -7, c = 3. The quadratic formula gives:
| | x | = (7 ± √(-7)² - 4·2·3 / 2·2 |
| | | = (7 ± √49 - 24 / 4 |
| | | = (7 ± √25 / 4 |
| | | = (7 ± 5) / 4 |
giving two roots, say, x1 = (7 + 5) / 4 = 3 and x2 = (7 - 5) / 4 = 1/2. It can also be verified that
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2x² - 7x + 3 = 2(x - 3)(x - 1/2).
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Obviously, the expression D = b² - 4ac that appears in the quadratic formula under the square root plays an important role in solving quadratic equations. Because of its importance it was given a name. D = b² - 4ac is called the determinant of the quadratic equation ax² + bx + c = 0. As the quadratic formula shows, there are three possible cases:
- D > 0. In this case, the equation has two distinct real roots.
- D = 0. In this case, the two roots coalesce into one, which is called a double root implying that the are still two roots albeit equal.
- D < 0. In this case, the equation has to complex roots which, for real a, b, c, are conjugate.
For quadratic polynomials, the roots come in pairs as manifest by the "±" symbol. For the polynomials with real coefficients but complex roots, the roots in such pairs are conjugate. (An interactive illustration of the behavior of the roots as the function of the coefficients is available elsewhere.) We now know that the same holds for all polynomials with real coefficients of degree higher than 1. This means that any polynomial with real coefficients admits a decomposition into a product of linear and quadratic terms. However, this fact was not always known of course. In the 17th century, this was still a conjecture. The Fundamental Theorem of Algebra is the product of the 19th century. According to [W. Dunham, p. 109],
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No less an authority than Leibniz doubted that every real polynomial can be factored into real linear and/or real quadratic pieces. Worse, Nicolaus Bernoulli (1687-1759) claimed to have found a counter example - namely, x4 - 4x3 + 2x2 + 4x + 4 - that could not be so factored. If he were correct, the game was over: the fundamental "theorem" of algebra would have been automatically disproved.
Euler, rising to the defence of this conjecture, showed that Bernoulli was wrong. In a 1742 letter to Christian Goldbach, he factored the supposedly unfactorable, splitting the quartic into the product of quadratics
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x² - (2 + √4 + 2√7)x + (1 + √4 + 2√7 + √7)
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and
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x² - (2 - √4 + 2√7)x + (1 - √4 + 2√7 + √7).
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This factorization appears to lie somewhere between miraculous and preposterous. It looks ever so much like a misprint - but it is perfectly correct.
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Every one is welcomed to check the verity of Euler's feat.
References
- A. De Morgan, On the Study and Difficulties of Mathematics, Dover, 2005, p. 149
- W. Dunham, Euler: The Master of Us All, MAA, 1999
Copyright © 1996-2009 Alexander Bogomolny
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