Questions and Answers #14 Quadratics

Questions and Answers Sheet 14 Quadratics Question #1 A little bit different quadratic Gauss sum Could anyone give me a direction on how to demonstrate that 𝜋 𝜋 2 −𝑖 (𝑘+𝐶) ∑𝑁−1 = √𝑁𝑒 −𝑖 4 , 𝑘=0 𝑒 𝑁 If 𝑁 ∈ 𝑁 is even, for any 𝐶 ∈ 𝑍. Unfortunately, the left side of the above equation cannot be exactly defined as a quadratic Gauss sum since it does not have the term 2 within the argument of e and the summation coefficient is shifted by C. Answer: Let 𝑁 = 2𝑛. Note the following: 4 n −1 − i 2 ( k +C )2 4n e k =0 4 n −1 − i 2 k 2 4n = e = (1 − i ) 4n = 2e −i  4 N k =0 2𝜋 Here we used that 𝑒 −𝑖4𝑛 is a primitive 4n-th root of unity and a classical Gauß sum identity (I used Lang's Algebraic Number Theory as a reference for that part). The former part is given since while k runs through a system of representatives of 𝑍/4𝑛 𝑍 so does k+C for all 𝐶 ∈ 𝑍. Now it only remains to show that the sum on the LHS is twice the sum from 𝑘 = 0 to 𝑘 = 2𝑛 − 1. For that part notice that 2 ((2𝑛 + 𝑘) + 𝐶) ≡ (𝑘 + 𝐶)2 𝑚𝑜𝑑 4𝑛 which a short computation reveals. Therefore, 2𝜋 2 2𝜋 2 2𝜋 2 2𝜋 2 −𝑖 (𝑘+𝐶) −𝑖 (𝑘+𝐶) −𝑖 (𝑘+𝐶) −𝑖 (𝑘+𝐶) ∑4𝑛−1 = ∑2𝑛−1 + ∑4𝑛−1 = 2 ∑2𝑛−1 𝑘=0 𝑒 4𝑛 𝑘=0 𝑒 4𝑛 𝑘=2𝑛 𝑒 4𝑛 𝑘=0 𝑒 4𝑛 which concludes the proof (well, up to the knowledge of the "classical Gauß sum identity"). Question #2 A quadratic equation question The number of positive integral values of k for which (16𝑥 2 + 12𝑥 + 39) + 𝑘(9𝑥 2 − 2𝑥 + 11) is a perfect square is: a) two b) zero c) one d) none of these Answer: This is assuming it's "can be a perfect square for some x". Let's call your expression F(k,x). For any positive integer x, F(k,x) is of the form a(x) k + b(x) for some integers a(x),b(x) . If b(x) is a quadratic residue mod a(x), then there are infinitely many positive integers k that make F(k,x) a square. For example, 𝐹(𝑘, 1) = 18𝑘 + 67, and 67 ≡ (±7)2 𝑚𝑜𝑑 18. Then for any 𝑦 ≡ ±7 𝑚𝑜𝑑 18 with 𝑦 2 > 67, we can take k to be the integer (𝑦 2 − 67)/18. The positive integer k's thus obtained are 3,31,43,99,119,203,231,… . Question #3 A quadratic 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 has its roots in the interval [0,1], find the maximum value of (𝑎−𝑏)(2𝑎−𝑏) 𝑎(𝑎−𝑏+𝑐) Answer: 𝑏 𝑐 Since 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 has the same roots as 𝑥 2 + 𝑥 + = 0 for 𝑎 ≠ 0 , this suggests writing instead 𝑎 𝑎 (𝑎 − 𝑏)(2𝑎 − 𝑏) (1 − 𝑏/𝑎)(2 − 𝑏/𝑎) = 𝑎(𝑎 − 𝑏 + 𝑐) 1 − 𝑏/𝑎 + 𝑐/𝑎 and letting 𝑝 = 𝑏/𝑎, 𝑞 = 𝑐/𝑎 , for which we now seek to maximize 𝑓(𝑝, 𝑞) = (1−𝑝)(2−𝑝) . 1−𝑝+𝑞 But if 𝑟1 , 𝑟2 ∈ [0,1] are the roots, we also have (𝑥 − 𝑟1 )(𝑥 − 𝑟2 ) = 𝑥 2 + 𝑝𝑥 + 𝑞, hence 𝑝 = −(𝑟1 + 𝑟2 ), 𝑞 = 𝑟1 𝑟2 , after which we find 𝑟1 𝑟2 𝑓(−𝑟1 − 𝑟2 , 𝑟1 𝑟2 ) = 2 + + . 1 + 𝑟2 1 + 𝑟1 Question #4 A question defines 𝑓(𝑥) = (𝑎𝑥 + 𝑏)𝑒 𝑥 and states that it satisfies the equation 𝑓(𝑥) ∫0 = 𝑥 𝑥−𝑦 ′ (𝑦)𝑑𝑦 𝑒 𝑓 − (𝑥 2 − 𝑥 + 1)𝑒 𝑥 You're required to find a and b. Answer: Going by your method, we get, 𝑓 ′ (𝑥) = 𝑒 𝑥 (𝑎𝑥 + 𝑏 + 𝑎) 𝑥 ⇒ 𝑓(𝑥) = ∫0 𝑒 𝑥−𝑦 𝑒 𝑦 (𝑎𝑦 + 𝑏 + 𝑎)𝑑𝑦 − (𝑥 2 − 𝑥 + 1)𝑒 𝑥 𝑥 ⇒ 𝑓(𝑥) = 𝑒 𝑥 ∫0 (𝑎𝑦+𝑏+𝑎)𝑑𝑦 − (𝑥 2 − 𝑥 + 1)𝑒 𝑥 ⇒ 𝑓(𝑥) = 𝑒 𝑥 ( 𝑎𝑥 2 2 + (𝑏 + 𝑎)𝑥) − (𝑥 2 − 𝑥 + 1)𝑒 𝑥 𝑎 ⇒ (𝑎𝑥 + 𝑏)𝑒 𝑥 = 𝑒 𝑥 (( − 1) 𝑥 2 + (𝑏 + 𝑎 + 1)𝑥 − 1) 2 𝑥 Cancelling 𝑒 and equating coefficient both sides, you get a=2 and b=-1 . 𝑥 Going by their method, we get, 𝑒 −𝑥 𝑓(𝑥) = ∫0 𝑒 −𝑦 𝑓 ′ (𝑦)𝑑𝑦 − (𝑥 2 − 𝑥 + 1) ⇒ 𝑑 𝑑𝑥 (𝑎𝑥 + 𝑏) = 𝑒 −𝑥 𝑓 ′ (𝑥) − (2𝑥 − 1) ⇒ 𝑎 = (𝑎𝑥 + 𝑏 + 𝑎) − (2𝑥 − 1) ⇒ 0 = (𝑎 − 2)𝑥 + (𝑏 + 1) = 0 which leads to the same solution. Note that the given answer that b=1 is wrong. Question #5 Zeros of polynomial 𝑓(𝑧) = 𝑧 2 + 𝑎𝑧 + 𝑝11 1) f has a zero on imaginary axis. 2) f has a zero for which real and imaginary parts are equal. 3) f has distinct roots. 4) f has exactly one root. Answer: Use Quadratic equation formula for the root of the polynomial. ∴𝑧= −𝑎±√𝑎2 −4𝑝11 2 Now Option (1), (2) and (4) can be eliminated by using 𝑎 = 4𝑝11 Question #6 You are given a quadratic function 𝑓(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 and a linear function g(x). The two functions intersect at x=0 and at also at an x with g(x)=f(x)=0 and where x<0. Which of the two could, for some values of a, b, c, be an expression for g(x): 1) g(x)=bx+c 2) g(x)=ax+c Answer: Given: 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐(𝑎 ≠ 0) If 𝑔(𝑥) = 𝑏𝑥 + 𝑐 for some 𝑥 = 𝑥1 < 0 𝑓(𝑥1 ) = 𝑎𝑥12 + 𝑏𝑥1 + 𝑐(1) 𝑔(𝑥1 ) = 𝑏𝑥1 + 𝑐(2) (1)-(2) 𝑎𝑥12 = 0, 𝑎 = 0, it is not allowed So, this is not the right choice, only option is (2) 𝑔(𝑥) = 𝑎𝑥 + 𝑐. Question #7 Would an equation in the form ax+b=c/x be considered quadratic? For example is 2𝑥 + 3 = 5 𝑥 quadratic? On the one hand it has two solutions, 𝑥 = 1 and 𝑥 = − 5 2 which is the number of solutions we'd expect from the fundamental theorem of algebra but on the other hand any equation in the form 𝑎𝑥 + 𝑏 = 𝑐 𝑥 would be undefined at x=0 and every quadratic equation I'm familiar with is continuous over all real numbers. Is being continuous over all real numbers necessary for an equation to be called quadratic? Answer: The fact that it has two solutions is not relevant here. The equation 2𝑥 − 𝑥 = 0 also has two solutions, but nobody would say that it is a quadratic equation. And the equation 𝑥 2 + 1 = 0 is a quadratic equation, in spite of the fact that it has no (real) solution. A reasonable definition of quadratic equation would be: an equation of the type 𝑞(𝑥) = 0, where q is a polynomial function with degree 2. It would still be a quadratic equation even if we were only interested in solutions within a certain subset of 𝑅 (such as, say, (−∞, 1] or 𝑄 ). Under this definition, your equation is not a quadratic equation. However, its solutions are the solutions of the quadratic equation 2𝑥 2 + 3𝑥 − 5 = 0. Question #8 Perform the multiplication or division and simplify. 𝑥+3 4𝑥 2 −9 ÷ 𝑥 2 +7𝑥+12 22 +7𝑥−15 Answer: simplify: 𝑥+3 4𝑥 2 −9 ÷ 𝑥 2 +7𝑥+12 22 +7𝑥−15 Inverting and changing to multiplication: 𝑥+3 4𝑥 2 −9 = ÷ 𝑥+3 4𝑥 2 −9 𝑥 2 +7𝑥+12 22 +7𝑥−15 𝑥 2 +7𝑥+12 × 22 +7𝑥−15 factorizing the quadratics: 𝑥+3 = (2𝑥)2 −32 × 2𝑥 2 +10𝑥−15 𝑥+3 𝑥 2 +4𝑥+12 2𝑥(𝑥+5)−3(𝑥+5) 𝑥+3 𝑥(𝑥 + 4)+3(𝑥+4) (2𝑥−3)(𝑥+5) = (2𝑥−3)(2𝑥+3) × = (2𝑥−3)(2𝑥+3) × (𝑥+3)(𝑥+4) cancelling the same terms, we get: = ( x + 3) (2 x − 3) (2 x + 3)  (2 x − 3) ( x + 5) ( x + 3) ( x + 4) (𝑥+5) = (2𝑥+3)(𝑥+4) Question #9 1. Looking around in your own country, discuss two practical applications in which “piecewise functions” are used either explicitly or implicitly. 2. To the best of your ability, using mathematical notion or in words, create any one of the functions you have spoken of in part (1) above. 3. If you could only learn one approach to solving quadratic equations, discuss which of the approaches you would choose and why? 4. Do you wholly agree with the statement that “the domain and range of a quadratic equation will be [−∝, ∝]”? Provide arguments to support your position Answer: Consider the provided question, (1) Piecewise functions are the kind of functions, which behaves differently in different domains. The examples of Piecewise functions are: (a) The Income tax in a bracketed system. As the slab changes, a sudden change appears in the function. (b) Cab fares in different long runs: The Ratio/slope of fair decreases for very long runs. while for short distance travelling, they are higher. (2) Let x be the "Taxable income over " and y be the "Rate". Then 10%, x  $9000  y = f ( x) = 15% $9000  x  $35, 000 20% x  $35, 000  (3) For solving quadratics, The simplest among all approach is "Shridharacharya Method" (quadratic formula). Because this is straightforward and easy. also, Real and Non real roots both can be found with equal ease. Shridharacharya Method says, For an quadratic equation like 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 we can find the value of x as, 𝑥 = −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 4) No, I don't agree with the given statement. Because any quadratic can be transformed as a sum of perfect square and a constant. While, Perfect square is always Non-Negative. Therefore, Range of Quadratic can't be [−∝, ∝]. Question #10 The complex zeros of 𝑓(𝑥) = 𝑥 4 + 1. For the function 𝑓(𝑥) = 𝑥 4 + 1: (a) Factor f into the product of two irreducible quadratics. (Hint: Complete the square by adding and subtracting 2𝑥 2 .) (b) Find the zeros of f by finding the zeros of each irreducible quadratic. Answer: a. 𝑓(𝑥) = 𝑥 4 + 2𝑥 2 + 1 − 2𝑥 2 2 = (𝑥 2 + 1)2 − (√2𝑥) Using (𝑎2 − 𝑏 2 ) = (𝑎 + 𝑏)(𝑎 − 𝑏) f ( x) = ( x2 + 1 + 2 x )( x2 + 1 − 2 x )