IX. REFLEXIONES FINALES
9.2 Reflexiones desde el trabajo social
1. In the first equation, since(2003)2−2002×2004−1 = 0, sox−1is a factor of the left hand side. By cross multiplication, it is obtained that
(x−1)(20032x+ 1) = 0, so the other root is− 1
20032, and the larger rootmis1.
For the second equation, Since1 + 2002−2003 = 0,x−1is a factor of the left hand side, so it follows that
(x−1)(x+ 2003) = 0,
the smaller rootnis−2003. Thusm−n= 1 + 2003 = 2004.
2. The partition points of the range ofxis−3and3. (i) Whenx≤ −3, thenx2−2x−24 = 0, sox
Lecture Notes on Mathematical Olympiad 141
(ii) When−3 < x ≤ 3, thenx2−18 = 0, sox = ±3√2 (Both are N.A.).
(iii) When3< x, thenx2+ 2x−24 = 0, sox= 4(x=−6is N.A.). Thus, the roots are−4,4.
3. Whenm= 2, the equation becomes−5x−11 = 0, the solution isx=−1. Whenm6= 2,∆ = [−(m+ 3)]2+ 4(2m+ 1)(m−2) = 9m2−6m+ 1 = (3m−1)2≥0, so x1= (m+ 3)−(3m−1) 2(m−2) =−1, x2= (m+ 3) + (3m−1) 2(m−2) = 2m+ 1 m−2 .
4. a2−3a+ 1 = 0yieldsa2+ 1 = 3aanda6= 0. Therefore by division,
2a5−5a4+ 2a3−8a2 a2+ 1 = a2−3a+ 1)(2a3+a2+ 3a)−3a a2+ 1 =−3a 3a =−1.
5. Letx0be the common root of the two given equations, then
1988x2
0+bx0+ 8891 = 8891x20+bx0+ 1988,
(8891−1988)x2
0= (8891−1988),
∴x0=±1. Substituting back suchx0into the first equation,
±b=−(8891 + 1988) =−10879, ∴b=∓10879.
6. (i) Whenm= 1, then−6x+ 1 = 0, so there is a real rootx= 1 6.
(ii) Whenm=−1, then−2x+ 1 = 0, so there is a real rootx= 1 2.
(iii) Whenm2−16= 1, then∆ = 4(m+2)2−4(m2−1) = 16m+20≥0 implies thatm≥ −5
4 andm6= 1.
Thus, the range ofmism≥ −5 4.
7. Letx0be the common root, thenx20−kx0−7 = 0andx20−6x0−(k+1) = 0. Then their difference gives
142 Solutions to Testing Questions
Notice thatk6= 6. Otherwise, the two equations are identical so that they have two common roots. Thereforex0= 1, and it implies thatk=−6. By substituting back the value ofkinto the given equations, it follows that
x2+ 6x−7 = 0 and x2−6x+ 5 = 0,
∴(x−1)(x+ 7) = 0 and (x−1)(x−5) = 0. Thus, the different roots are−7and5respectively.
8. Since the given equation has two equal real roots implies that its discriminant is0, so
∆ = (2b)2−4(c+a)(c−a) = 4(b2+a2−c2) = 0,
∴a2+b2=c2.
Fromc2 = a2+b2 < a2+b2+ 2ab = (a+b)2, it follows thatc < a+b, so the three segments with lengthsa, b, ccan form a triangle. Further, from Pythagoras’ Theorem, the triangle is a right-angled triangle with the hypotenuse side of lengthc.
9. Since the discriminant of the equation is non-negative,
4(1 +a)2−4(3a2+ 4ab+ 4b2+ 2)≥0, (1 +a)2−(3a2+ 4ab+ 4b2+ 2)≥0, 2a2−2a+ 1 + 4ab+ 4b2≤0, (a−1)2+ (a+ 2b)2≤0, ∴a= 1, and a+ 2b= 0, a= 1, b=−1 2. 10. The discriminant of the equation is given by
∆ = (a+b+c)2−4(a2+b2+c2)
= −3a2−3b2−3c2+ 2ab+ 2bc+ 2ca
= −(a2−2ab+b2)−(b2−2bc+c2)
−(c2−2ca+a2)−(a2+b2+c2)
= −[(a−b)2+ (b−c)2+ (c−a)2+ (a2+b2+c2)]<0. so the equation has no real roots, the answer is (D).
Lecture Notes on Mathematical Olympiad 143
Testing Questions (24-B)
1. Mr. Fat has the winning strategy. A set of three distinct rational nonzero numbersa, b, andc, such thata+b+c= 0, will do the trick. LetA, B, and Cbe any arrangement ofa, b, andc, and letf(x) =Ax2+Bx+C. Then
f(1) =A+B+C=a+b+c= 0, which implies that1is a solution.
Since the product of the two solutions isC
A, the other solution is C A, and it is different from1.
2. By∆i, i= 1,2,3we denote the discriminant of theith equation. It suffices to show∆1+ ∆2+ ∆3>0. Then
∆1+ ∆2+ ∆3 = (4b2−4ac) + (4c2−4ab) + (4a2−4bc)
= 2(2a2+ 2b2+ 2c2−2ab−2bc−2ca)
= 2[(a−b)2+ (b−c)2+ (c−a)2]>0. Thus, the conclusion is proven.
3. From the assumptionsa >1andb >1anda 6=b. Letx0be the common root, then
(a−1)x20−(a2+2)x0+(a2+2a) = 0,(b−1)x20−(b2+2)x0+(b2+2b) = 0. Notice thatx0 6= 1. Otherwise, ifx0 = 1, then above two equalities be- comesa= 1 =b.
After eliminating the term ofx2
0and doing simplification, it follows that
(a−b)(ab−a−b−2)(x0−1) = 0.
Sincea−b6= 0andx06= 1, soab−a−b−2 = 0, i.e.ab=a+b+ 2. (i) Whena > b >1, thenb= 1 +b
a+
2
a <3, sob= 2, a=
4
b−1 = 4.
(ii) Whenb > a >1, then, by symmetry,a= 2, b= 4. Thus, in each of above two cases,
ab+ba a−b+b−a = (a
b+ba)· abba ab+ba =a
144 Solutions to Testing Questions
4. Since the left hand side cannot be zero, som6= 0andm >0. (i) Whenm >0, thenx >0.
For0< x≤1, then1−x2+ 4−x2=mx, i.e.2x2+mx−5 = 0, so
x=−m+
√
m2+ 40
4 ≤1.
It implies thatm2+ 40≤m2+ 8m+ 16, i.e.3≤m. For1< x≤2, thenx2−1 + 4−x2=mx, so1< x= 3 m ≤2, i.e. 3 2 ≤m <3, for2< x, then2x2−mx−5 = 0, sox=m+ √ m2+ 40 4 >2, it implies that m2+ 40> m2−16m+ 64, i.e.m > 3 2.
(ii) Whenm <0, thenx <0, so the case can be converted to the case (i) if use−m,−xto replacem, xrespectively. Thus, the solutions are
x= 3 m or m+√m2+ 40 4 if 3 2 ≤m <3, −m+√m2+ 40 4 ifm≥3, 3 m or m−√m2+ 40 4 if−3< m≤ − 3 2, −m+√m2+ 40 4 ifm≤ −3,
and has no solution for other values ofm.
5. Suppose that the first equation has no two distinct real roots, then
∆1= 1−4q1≤0, i.e.q1≥1
4.
In this case, then the discriminant of the second equation is
∆2 = p2−4q2= (q1+q2+ 1)2−4q2
= q2
2+ 2q2(q1+ 1) + (q1+ 1)2−4q2
= q2
Lecture Notes on Mathematical Olympiad 145
To show∆2is always positive, consider the last expression as a quadratic function ofq2, then its discriminant is
∆3= 4(a1−1)2−4(1 +q1)2=−16q1≤ −4,
so∆2>0for any value ofq2, i.e. the second equation has two distinct real roots, the conclusion is proven.