Math IOQM

For a.x^2 + b.x + c = 0 to have rational roots , necessary and sufficient conditions are: 1. b/a and c/a must be rational. 2. Discriminant must be square of a rational number. Similarly for integer roots: 1. b/a and c/a must be integers. 2. Discriminant must be perfect square of an integer. Let's go in some more detail here. If b/a and c/a are integers I can rewrite the equation as: x^2 + B.x + C = 0 where B = b/a and C = c/a. roots = {-B +- sqrt(B^2 - 4C)}/2 Case 1:  B is even. roots = (even +- even)/2 = integer Case 2: B is odd. roots = (odd +- odd)/2 = integer So yeah, we are good. Ex1: Find possible values of k k such that x 2 − k x + 4 = 0 x^2 - kx + 4 = 0 has integer roots, given that k ∈ { − 10 , − 9 , − 8 , … , 9 , 10 } k \in \{-10, -9, -8, \ldots, 9, 10\} Solution 1: Discriminant >= 0 and Perfect square of an integer => k^2 - 16 = m^2 where m is an integer. (k+m).(k-m) = 16 Now LHS can take on factors of 16: 1,16 2,8 4,4 8,2 16,1 -1,-16 -2,-8 -4,-4 -8,-2 -16,-1...

Question 5: $$ \text{If } a^2 + 5b^2 + 4c^2 - 4ab - 4bc = 0 \text{ then find the value of } \frac{a^2 + ac}{b^2}. $$ Solution: 1. First we will complete the square in "a" since a^2 has the coefficient of 1. So we get: \((a -2b)^2 + ... = 0\) 2. Now we complete the square of the remaining terms to get: \((a - 2b)^2 + (b - 2c)^2 = 0.\) If two squares add up to 0 then they are individually 0. So a = 2b and b = 2c. Put the value of a,c in terms of b in the question to get the answer which is 5.

9. Suppose a , b a, b are integers and a + b a + b is a root of x 2 + a x + b = 0 x^2 + ax + b = 0 . What is the maximum possible value of b 2 b^2 ? [PRE-RMO – 2018] We are given that a + b a + b is a root of the quadratic equation: x 2 + a x + b = 0 x^2 + ax + b = 0 Substituting x = a + b x = a + b into the equation: ( a + b ) 2 + a ( a + b ) + b = 0 (a + b)^2 + a(a + b) + b = 0 Expanding the terms: a 2 + 2 a b + b 2 + a 2 + a b + b = 0 a^2 + 2ab + b^2 + a^2 + ab + b = 0 2 a 2 + 3 a b + b 2 + b = 0 2a^2 + 3ab + b^2 + b = 0 This is a quadratic equation in a a : 2 a 2 + 3 a b + ( b 2 + b ) = 0 2a^2 + 3ab + (b^2 + b) = 0 To ensure that real values of a a exist, the discriminant D D of this quadratic must be a perfect square. The discriminant is: D = ( 3 b ) 2 − 4 ⋅ 2 ⋅ ( b 2 + b ) = 9 b 2 − 8 ( b 2 + b ) D = (3b)^2 - 4 \cdot 2 \cdot (b^2 + b) = 9b^2 - 8(b^2 + b) D = 9 b 2 − 8 b 2 − 8 b = b 2 − 8 b D = 9b^2 - 8b^2 - 8b = b^2 - 8b Let D = p 2 D = p^2 , where p ∈ Z p ...

8. The equations x 2 − 4 x + k = 0 x^2 - 4x + k = 0 and x 2 + k x − 4 = 0 x^2 + kx - 4 = 0 , where k k is a real number, have exactly one common root. What is the value of k k ? [PRE-RMO – 2015] Solution: Let the common root be 'b'. First equation has roots a,b and the second one has b,c. a + b = 4 ab = k a = 4-b = k/b => 4b - b^2 = k => b^2 = 4b - k ____[1] Similarly, b + c = -k bc = -4 c = -k-b = -4/b => b^2 + kb = 4 => b^2 = 4 - kb________[2] From [1] and [2] 4 - kb = 4b - k => b = 1 => k = 3 Answer: 3 Solution 2: Suppose the two quadratics have a common root α \alpha . Then { α 2 − 4 α + k = 0 , α 2 + k α − 4 = 0. \begin{cases} \alpha^2 - 4\alpha + k = 0, \\ \alpha^2 + k\alpha - 4 = 0. \end{cases} Subtracting one equation from the other gives ( α 2 + k α − 4 )    −    ( α 2 − 4 α + k )    =    ( k + 4 )   α    −    ( k + 4 )    =    ( k + 4 ) ( α − 1 ) = 0. (\alpha^2 + k\alpha - 4) \;-\; (\alpha^2 - 4\alpha + k) \;=\; (k + 4)\,\alpha \;-\; (k+4...

Question: What is the the maximum possible value of k for which 2013 can be written as sum of k consecutive positive integers? Solution: Suppose that we start with some a ∈ N a \in \mathbb{N} . Then we want that a + ( a + 1 ) + … + ( a + ( k − 1 ) ) = 2013. a + (a+1) + \ldots + (a+(k-1)) = 2013. Rewriting it as the sum of arithmetic series we get ( 2 a + k − 1 ) k 2 = 2013 or ( 2 a − 1 + k ) k = 4026. \frac{(2a + k - 1)k}{2} = 2013 \quad \text{or} \quad (2a - 1 + k)k = 4026. Note that 2 a − 1 + k ≥ k 2a - 1 + k \geq k . So k, (2a - 1 + k) are 2 factors of 4026 such that 'k' is the smaller factor. We need to find all factor pairs of 4026 and take the biggest smaller factor. Now we find factor pairs of 4026. The factor pairs of 4026 are: - (1, 4026) - (2, 2013) - (3, 1342) - (6, 671) - (11, 366) - (22, 183) - (33, 122) - (61, 66) The biggest smaller factor is 61. Answer: 61. 2a - 1 + k = 66 = 2a + 60 => a = 3 Cross check that the A.P. starting from 3 and having 61 terms...

4. Let  f ( x ) = x 3 − 3 x + b f(x) = x^3 - 3x + b and g ( x ) = x 2 + b x − 3 g(x) = x^2 + bx - 3 , where b b is a real number. What is the sum of all possible value of b b for which the equation f ( x ) = 0 f(x) = 0 and g ( x ) = 0 g(x) = 0 have a common root? [PRE-RMO – 2013] Solution: Suppose  x = α x=\alpha is a common root of f ( x )    =    x 3 − 3 x + b and g ( x )    =    x 2 + b x − 3. f(x) \;=\; x^{3}-3x+b \quad\text{and}\quad g(x) \;=\; x^{2}+bx-3. Then α 3 − 3 α + b    =    0 , \alpha^{3}-3\alpha + b \;=\; 0, α 2 + b   α − 3    =    0. \alpha^{2}+b\,\alpha - 3 \;=\; 0. From the second equation, provided α ≠ 0 \alpha \neq 0 , solve for b b : b    =    3 − α 2 α . b \;=\;\frac{3 - \alpha^{2}}{\alpha}. Substitute this into the first equation: α 3 − 3 α    +    3 − α 2 α    =    0 , \alpha^{3} - 3\alpha \;+\; \frac{3 - \alpha^{2}}{\alpha} \;=\; 0, and multiply through by α \alpha (again using α ≠ 0 \alpha \neq 0 ): α 4    −    3 α 2    +    ( 3  ...

3. It is given that the equation  x 2 + a x + 20 = 0 x^2 + ax + 20 = 0 has integer roots. What is the sum of all possible values of a a ? [PRE-RMO – 2013] Solution 1: x = (a ± √(a² - 80)) / 2 a² - 80 should be a perfect square. For eg. a = 9 ⇒ √1 a = -9 also works And x should be integer. So numerator, i.e. a ± √(a² - 80) should be even. If a is odd, (a² - 80) is odd and hence √(a² - 80) is odd ⇒ numerator = -a ± odd = -odd ± odd = even Similarly if a is even, (a² - 80) is even and hence √(a² - 80) is even ⇒ numerator = -a ± even = -even ± even = even So numerator is always even if (a² - 80) is a perfect square. And for each valid value of (a² - 80), there are 2 valid values of 'a'(plus and minus). So no matter how many valid values of 'a' are there, they will all add up to 0. Answer: 0 Solution 2: A convenient way to proceed is to let the integer roots be  r 1 r_1 and r 2 r_2 . Then by Vieta’s formulas, r 1 + r 2 = − a and r 1   r 2 = 20. r_1 + r_2 = -a \quad\te...

Let x 1 , x 2 , x 3 x_1, x_2, x_3 be the roots of the equation x 3 + 3 x + 5 = 0. x^3 + 3x + 5 = 0. What is the value of the expression ( x 1 − 1 x 1 ) ( x 2 − 1 x 2 ) ( x 3 − 1 x 3 ) \left( x_1 - \frac{1}{x_1} \right) \left( x_2 - \frac{1}{x_2} \right) \left( x_3 - \frac{1}{x_3} \right) [PRE-RMO – 2012] Prerequisites: Vieta's formulas And (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) Source Sol: x 3 + 0 x 2 + 3 x + 5 = 0 x^3 + 0x^2 + 3x + 5 = 0 have roots x 1 , x 2 , x 3 x_1, x_2, x_3 ⇒ x 1 + x 2 + x 3 = 0 ; x 1 x 2 + x 2 x 3 + x 3 x 1 = 3 ; x 1 x 2 x 3 = − 5 \Rightarrow x_1 + x_2 + x_3 = 0 ; \quad x_1x_2 + x_2x_3 + x_3x_1 = 3 ; \quad x_1x_2x_3 = -5 ( x 1 + 1 x 1 ) ( x 2 + 1 x 2 ) ( x 3 + 1 x 3 ) ⇒ ( x 1 x 2 + x 1 x 2 + x 2 x 1 + 1 x 1 x 2 ) ( x 3 + 1 x 3 ) (x_1 + \frac{1}{x_1}) (x_2 + \frac{1}{x_2}) (x_3 + \frac{1}{x_3}) \Rightarrow (x_1x_2 + \frac{x_1}{x_2} + \frac{x_2}{x_1} + \frac{1}{x_1x_2})(x_3 + \frac{1}{x_3}) = ( x 1 x 2 x 3 + x 1 x 2 x 3 + x 3 x 1 x 2...

Answer: 1