When b 2 ac the equation has only one root for the

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When b 2 = ac the equation has only one root. For the sake of uniformity it is generally said in this case to have ‘two equal’ roots, but this is a mere convention.
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[I : 14] REAL VARIABLES 23 Suppose a , b , c rational. Nothing is lost by taking all three to be integers, for we can multiply the equation by the least common multiple of their denominators. The reader will remember that the roots are {- b ± b 2 - ac } /a . It is easy to construct these lengths geometrically, first constructing b 2 - ac . A much more elegant, though less straightforward, construction is the following. Draw a circle of unit radius, a diameter PQ , and the tangents at the ends of the diameters. Q Q X Y P P N M Fig. 5. Take PP 0 = - 2 a/b and QQ 0 = - c/ 2 b , having regard to sign. * Join P 0 Q 0 , cutting the circle in M and N . Draw PM and PN , cutting QQ 0 in X and Y . Then QX and QY are the roots of the equation with their proper signs. The proof is simple and we leave it as an exercise to the reader. Another, perhaps even simpler, construction is the following. Take a line AB of unit length. Draw BC = - 2 b/a perpendicular to AB , and CD = c/a perpendicular to BC and in the same direction as BA . On AD as diameter describe a circle cutting BC in X and Y . Then BX and BY are the roots. 3. If ac is positive PP 0 and QQ 0 will be drawn in the same direction. Verify that P 0 Q 0 will not meet the circle if b 2 < ac , while if b 2 = ac it will be a tangent. Verify also that if b 2 = ac the circle in the second construction will touch BC . 4. Prove that pq = p × q, p p 2 q = p q. * The figure is drawn to suit the case in which b and c have the same and a the opposite sign. The reader should draw figures for other cases. I have taken this construction from Klein’s Le¸ cons sur certaines questions de eom´ etrie ´ el´ ementaire (French translation by J. Griess, Paris, 1896).
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[I : 14] REAL VARIABLES 24 14. Some theorems concerning quadratic surds. Two pure quadratic surds are said to be similar if they can be expressed as rational multiples of the same surd, and otherwise to be dissimilar . Thus 8 = 2 2 , q 25 2 = 5 2 2 , and so 8, q 25 2 are similar surds. On the other hand, if M and N are integers which have no common factor, and neither of which is a perfect square, M and N are dissimilar surds. For suppose, if possible, M = p q r t u , N = r s r t u , where all the letters denote integers. Then MN is evidently rational, and therefore ( Ex. ii . 3) integral. Thus MN = P 2 , where P is an integer. Let a , b , c, . . . be the prime factors of P , so that MN = a 2 α b 2 β c 2 γ . . . , where α , β , γ, . . . are positive integers. Then MN is divisible by a 2 α , and therefore either (1) M is divisible by a 2 α , or (2) N is divisible by a 2 α , or (3) M and N are both divisible by a . The last case may be ruled out, since M and N have no common factor. This argument may be applied to each of the factors a 2 α , b 2 β , c 2 γ , . . . , so that M must be divisible by some of these factors and N by the remainder. Thus M = P 2 1 , N = P 2 2 , where P 2 1 denotes the product of some of the factors a 2 α , b 2 β , c 2 γ , . . . and P 2 2 the product of the rest.
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