Questions and Answers #58 Quadratic Function and Equation

Questions and Answers Sheet 58 Quadratic Function and Equation Question #1 Let x and y be two real positive integers, such that: 𝑥 + 𝑦 + 𝑥𝑦 = 3 prove that 𝑥 + 𝑦 ≥ 2 . Answer: 3 = 𝑥 + 𝑦 + 𝑥𝑦 ≤ 𝑥 + 𝑦 + ( 𝑥+𝑦 2 2 ) , which gives (𝑥 + 𝑦)2 + 4(𝑥 + 𝑦) ≥ 12 or (𝑥 + 𝑦 + 2)2 ≥ 16 or 𝑥 + 𝑦 ≥ 2. Question #2 Let x be an integer and let d be a factor of 2𝑥 2 − 1. Prove that d  1mod 16 2 Answer: If d is a factor of 2𝑥 2 − 1 , then any ' factor p of d is a factor of 2𝑥 2 − 1 . Prime 𝑝 > 2 is a factor of 2𝑥 2 − 1 ⟺ 2 is a quadratic residue modulo p. p  1mod 8. Obviously 2 is not a factor of 2𝑥 2 − 1 . So d is a factor of 2𝑥 2 − 1 ⟺ 𝑑 ≡ ±1 mod 8 ⟺ 𝑑 2 ≡ 1 mod16 The last equivalence follows because if 8|𝑑 ± 1 then 16 | (d + 1)(d − 1) = d − 1 . 2 Question #3 Line 𝑥 + 𝑘𝑦 + 𝑘 2 = 0 is tangent to curve 𝑦 2 = 4𝑥 at point P. Find the coordinates of P in terms of k. Answer: Step 1 We are told that 𝑥 = −𝑘𝑦 − 𝑘 2 is tangential to 1 𝑥 = 𝑦2 4 We have 𝑑𝑥 𝑑𝑦 2 = 𝑦 4 2 4 𝑦 = −𝑘 1 Hence 𝑦 = −2𝑘 and 𝑥 = (−2𝑘)2 = 𝑘 2 2 Question #4 Linear matrix inequality given the bounds If 𝐴 ⪰ 𝐴𝑚𝑖𝑛 ≻ 0, 𝐵 ⪰ 𝐵𝑚𝑖𝑛 ≻ 0l the following be true? 𝐴𝐵𝐴⊤ ⪰ 𝐴𝐵𝑚𝑖𝑛 𝐴⊤ How can I prove it? Answer: We can prove by using definition of positive-definiteness. For any vector v  N and v  0 , it's obvious that from the definition 𝑣 ⊤ (𝐵 − 𝐵𝑚𝑖𝑛 )𝑣 ≥ 0 Since 𝐴 ⪰ 0 , for any vector x  and 𝑥 ≠ 0 , we have 𝐴⊤ 𝑥 ≠ 0 because A has full rank. Thus 𝑥 ⊤ 𝐴(𝐵 − 𝐵𝑚𝑖𝑛 )𝐴⊤ 𝑥 > 0 Therefore 𝐴(𝐵 − 𝐵𝑚𝑖𝑛 )𝐴⊤ ⪰ 0 𝑎nd 𝐴𝐵𝐴⊤ ⪰ 𝐴𝐵𝑚𝑖𝑛 𝐴⊤ N Question #5 Local extrema of the function 𝑓(𝑥) = 𝑥 3 − 3𝛼𝑥 2 + 3(𝛼 2 − 1)𝑥 + 1 Answer: Note that f ( x) = 3x 2 − 6 x + 3( 2 − 1) = 3[ x 2 − 2ax + (a + 1)(a − 1)] = 3[ x − (a + 1)][ x − (a − 1)] = 0 which leads to the roots 𝑥 = 𝑎 ± 1 . Then, solve −2 < 𝑎 ± 1 < 4 to obtain 𝑎 ∈ (−1,3). Question #6 Making a Negative Number Possible to Square Root We are able to solve 𝑥 2 + 4 = 0 by square rooting both sides, but if we have 𝑥 2 = −4 we can't solve. Firstly, why? Aren't they equal expressions? Secondly, if we have 𝑥 2 = −4 , why can't we bring the four to the other side, square root it and then bring it back? Answer: Actually there are a few assumptions we need to explicit. First of all we need to explicit where we are searching the solution for our equation. We can search for natural, integer, rational, real, complex solutions. In this case your question involves the real case. Let's analize the real case. We can easily see that the problem has no solution because 𝑥 2 ≥ 0 and so 𝑥 2 + 4 ≥ 4 . We can take the square root of both sides of the equation 𝑥 2 + 4 = 0 . Before taking the square root we must assume that each side is nonnegative. This is true. Now we have the equation √𝑥 2 + 4 = 0 . Again this equation has no solution. If you consider the equation 𝑥 2 = −4 , you cannot take the square root of both sides, because one of them is nonnegative. Doing this you are going outside the real numbers and this bring us to the next case. When you are taking a square root in the real numbers you are always assuming that what will be under the root is nonnegative Question #7 Making a Quadratic Function spit out Integer Outputs for Rational Inputs Suppose I have a quadratic function f ( x) = ax 2 + bx + c, a, b, c  and I want to find rational (non-integer) values of x for which 𝑓(𝑥) ∈ 𝑍 . Answer: You need to add the condition that 𝑞 ≠ ±1 . (I would define p,q as integers satisfying gcd(𝑝, 𝑞) = 1, 𝑞 ≥ 0 and 𝑞 ≠ 1 .) Except this, what you've done is correct. If there are integers p,q,n,k such that 𝑎 = 𝑛𝑞 and 𝑏 = 𝑘𝑞 − 𝑛𝑝 , then 𝑝 𝑝 2 𝑝 𝑞 𝑞 𝑞 𝑓 ( ) = 𝑛𝑞 ( ) + (𝑘𝑞 − 𝑛𝑝) ⋅ + 𝑐 = 𝑘𝑝 + 𝑐 is an integer. Question #8 Making sense of William Jones's solution of quadratic equations and notation William Jones, when discussing quadratic equations, says: "Therefore if ∨ be put for the Sign of any Term, and ∧ for the contrary, all Forms of Quadratics with their Solutions, will be reduc'd to this one. If 1 1 𝑥 𝑥 ∨ 𝑎 𝑥 ∨ 𝑏 = 0 then  a  aa  b| 2 ." 2 Could you please help me make sense: (a) What is meant with "the contrary" of "any sign"? (b) Do I understand correctly that the "then" part supposed to be a solution to the preceding quadratic equation? If yes, how does this work exactly? How does it align with how we would symbolize the solution today? Answer: By "contrary", it looks like he just means "opposite sign" - so if ∨ is "+" then ∧ is "-" and vice versa. 1 In other words, this is saying that the solutions to 𝑥 2 + 𝑎𝑥 + 𝑏 = 0 are − 𝑎 + √𝑎2 − 𝑏 and similarly the 1 2 solutions to 𝑥 2 − 𝑎𝑥 − 𝑏 = 0 are + 𝑎 + √𝑎2 + 𝑏 , although it looks like there's a factor missing 2 somewhere that would actually make the expression correct (there might also be some grouping involved in the way the notation works that I haven't picked up on). Question #9 Manipulation of equation into quadratic form 𝑐√𝑑1 +𝑑2 −√𝑑1 𝑧1 −𝛿(𝑑2 )√𝑃0 𝑃1 √𝑑1 +𝑑2 = −𝑧𝛽 2 2 and (𝛿 2 𝑝1 𝑝0 )𝑑22 − 𝑑2 [(𝑐 + 𝑧𝛽 ) − 2√𝑑1 𝑝1 𝑝0 𝑧1 𝛿] + 𝑑1 [𝑧12 − (𝑐 + 𝑧𝛽 ) ] = 0 Answer: At first glance, it seems that the difficulty lies in the combination of terms that contain 𝑑2 and √𝑑1 + 𝑑2 in the numerator, because if you simply square the expression, the two factors will mix, and you won't be able to get rid of the square root over √𝑑1 + 𝑑2 in the mix term. However, note that the denominator itself is √𝑑1 + 𝑑2 , so you can split off the term 𝑐√𝑑1 + 𝑑2 in the numerator, and then the left hand side becomes, 𝑐√𝑑1 +𝑑2 −𝑧1√𝑑 −𝑑2 𝛿√𝑝1 𝑝0 1 √𝑑1 +𝑑2 = 𝑐√𝑑1 +𝑑2 √𝑑1 +𝑑2 − 𝑧1√𝑑 +𝑑2 𝛿√𝑝1 𝑝0 1 √𝑑1 +𝑑2 . This allows you to rearrange the equation into z1 d1 + d 2 p1 p0 d1 + d 2 = c + z and further into 𝑧1√𝑑1 + 𝑑2 𝛿 √𝑝1 𝑝0 = (𝑐 + 𝑧𝛽 )√𝑑1 + 𝑑2 . Now you can square both sides and rearrange it into a quadratic equation in 𝑑2 . Question #10 Math Olympiad question: If the equation in x has real roots, then find the value of a and b. 𝑥 2 + 2(1 + 𝑎)𝑥 + (3𝑎2 + 4𝑎𝑏 + 4𝑏 2 + 2) = 0 Answer: For at least one real root: 𝑏 2 − 4𝑎𝑐 ≥ 0 So: [2(1 + 𝑎)]2 − 4(3𝑎2 + 4𝑎𝑏 + 4𝑏 2 + 2)(1) ≥ 0 1 + 𝑎2 + 2𝑎 − 3𝑎2 − 4𝑎𝑏 − 4𝑏 2 − 2 ≥ 0 𝑎2 + 2𝑎𝑏 + 2𝑏 2 − 𝑎 + 1 ≤ 0 (𝑎 + 𝑏)2 + 𝑏 2 − 𝑎 + 1 ≤ 0