Questions and Answers #61 Quadratic Function and Equation

Questions and Answers Sheet 61 Quadratic Function and Equation Question #1 Prove quadratic equation (𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐) has only one line of symmetry Answer: Step 1 Symmetry around 𝑥 = 𝑠 is expressed by 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 𝑎(2𝑠 − 𝑥)2 + 𝑏(2𝑠 − 𝑥) + 𝑐 for all x. By identification of the powers of x, we get the compatibility condition 𝑏 + 2𝑎𝑠 = 0 or 𝑠=− 𝑏 2𝑎 Step 2 Consider an arbitrary parabola and an arbitrary line. By rotation we can bring the line on the axis y. Then the equation of the rotated parabola 𝑎𝑥 2 + 2𝑏𝑥𝑦 + 𝑐𝑦 2 + 2𝑑𝑥 + 2𝑒𝑦 + 𝑓 = 0 must be invariant to a change of the sign of x and it must reduce to 𝑎𝑥 2 + 𝑐𝑦 2 + 2𝑒𝑦 + 𝑓 = 0. For this equation to represent a parabola, we must have 𝑐 = 0 . Then, 𝑎𝑥 2 + 2𝑒𝑦 + 𝑓 = 0 is a parabola for which the axis of symmetry is y. Hence any line which is an axis of symmetry is the standard axis. Step 3 Note that in the case 𝑎𝑐 ≠ 0 (centered conic), by translation we can further reduce to 𝑎𝑥 2 + 𝑐𝑦 2 + 𝑓 = 0 and the line is one of the known axis of symmetry. Hence there are exactly two of them. Question #2 Prove that 𝜋𝑒 𝑥−𝑒 + 𝑒𝜋 𝑥−𝜋 + 𝜋𝜋 +𝑒 𝑒 𝑥−𝜋−𝑒 = 0 has one real root in (𝑒, 𝜋) and other in (𝜋, 𝜋 + 𝑒). Answer: The roots of 𝑓(𝑥) = 𝜋𝑒 𝑒𝜋 𝜋𝜋 + 𝑒𝑒 + + =0 𝑥−𝑒 𝑥−𝜋 𝑥−𝜋−𝑒 and 𝑔(𝑥) = 𝜋 𝑒 (𝑥 − 𝜋)(𝑥 − 𝜋 − 𝑒) + 𝑒 𝜋 (𝑥 − 𝑒)(𝑥 − 𝑒 − 𝜋) + (𝜋 𝜋 + 𝑒 𝑒 )(𝑥 − 𝑒)(𝑥 − 𝜋) = 0 would coincide Note that 𝑔(𝑒) = 𝜋 1+𝑒 (𝜋 − 𝑒) > 0, ≈ 𝑔(𝜋) = 𝑒 𝜋+1 (𝜋 − 𝑒) < 0, ≈ 𝑔(𝑒 + 𝜋) = 𝑒𝜋(𝜋 𝜋 + 𝑒 𝑒 ) > 0 This proves that the quadratic g(x) has one real root in (𝑒, 𝜋) and other one in (𝜋, 𝜋 +)𝑒 . So will be the case for 𝑓(𝑥) = 0. Question #3 Prove that 12(𝑎𝑏 + 𝑏𝑎 + 𝑎𝑐) < 7𝑎2 + 15𝑏 2 + 18𝑐 2 holds for all positive numbers. Answer: Step 1 By AM-GM, or completing the square, (𝑝 − 𝑞)2 ≥ 0 ⇒ 𝑝2 + 𝑞 2 ≥ 2𝑝𝑞 we have (4a 2 + 9b 2 ) + (3a 2 + 12c 2 ) + (6b2 + 6c 2 )  12ab + 12ca + 12bc with equality for 2𝑎 = 3𝑏, 𝑎 = 2𝑐, 𝑏 = 𝑐 whose simultaneous solution is (𝑎, 𝑏, 𝑐) = (0,0,0) For 𝑎, 𝑏, 𝑐 > 0 , we have strict inequality. Question #4 Prove that 𝑓(𝑓(𝑥)) ≥ 0 for all real x Answer: Step 1 As you mention, 𝑎 ≥ 1 4 1 works. The only case that remains is when 0 < 𝑎 < , i.e. when f(x) has two 4 distinct real roots. In this case, it is enough to ensure that the midpoint of the roots, i.e. 𝑥 = − in a non-negative value, so we check 𝑓 (− 1 1 1 1 )= − +1=1− 2𝑎 4𝑎 2𝑎 4𝑎  f ( f (− 21a )) = f (1 − 41a ) = 16a 2 + 24a − 3 16a and from 16𝑎2 + 24𝑎 − 3 0 we get 𝑎 ≥ 2√3−3 4 1 2𝑎 , results Question #5 Prove that two coefficients of a quadratic function are always positive Let f(x) be a quadratic function of the form 𝑥 2 + 𝑝𝑥 + 𝑞 such that the equation 𝑓(𝑓(𝑥)) = 0 has two equal real roots. Show that 𝑝 ≥ 0 and 𝑞 ≥ 0 . Answer: Since 𝑓(𝑓(𝑥)) = 0 does have a solution, f(x) can be factored, into say 𝑓(𝑥) = (𝑥 − 𝛼)(𝑥 − 𝛽) . Here, we have 𝛼 + 𝛽 = −𝑝, 𝛼𝛽 = 𝑞. If 𝛼 = 𝛽 , you can show 𝛼 and must be 0. In this case, we have 𝑝 = 𝑞 = 0 . Now, assume 𝛼 ≠ 𝛽. Then, 𝑓(𝑓(𝑥)) = (𝑓(𝑥) − 𝛼)(𝑓(𝑥) − 𝛽) = 0 is equivalent to 𝑓(𝑥) − 𝛼 = 0 or 𝑓(𝑥) − 𝛽 = 0 . Since both of these equations cannot be true at the same time (since 𝛼 ≠ 𝛽 ) and 𝑓(𝑓(𝑥)) = 0 has only one solution, it must be the case that one of the two equations has only one root (i.e., a double root), and the other no roots. Without loss of generality, we can assume 𝑓(𝑥) − 𝛼 = 0 has a double root, and 𝑓(𝑥) − 𝛽 = 0 has no roots. Looking at the discriminant, we get (−𝛼 − 𝛽)2 − 4(𝛼𝛽 − 𝛼) = 0 and (−𝛼 − 𝛽)2 − 4(𝛼𝛽 − 𝛽) < 0. Now, we have (−𝛼 − 𝛽)2 − 4(𝛼𝛽 − 𝛼) = (𝛼 − 𝛽)2 + 4𝛼 = 0 , so 𝛼 ≤ 0 . Similarly, we have 𝛽 < 0 . Hence, 𝑝 = −𝛼 − 𝛽 ≥ 0 , and 𝑞 = 𝛼𝛽 ≥ 0 Question #6 Prove the following for the given quadratic If the difference of the roots of 𝑥 2 − 𝑝𝑥 + 𝑞 = 0is unity then prove that 𝑝2 − 4𝑞 = 1and𝑝2 = 4𝑞 2 = (1 + 2𝑞)2 Answer: You cannot prove it because second part is wrong. For example, the roots of 𝑥 2 − 𝑥 = 0 are 0 and 1 so the hypothesis is satisfied. We do have 𝑝2 − 4𝑞 = 1 but we do not have 𝑝2 = 4𝑞 2 = (1 + 2𝑞)2 . Actually, the difference between the roots is 1 if and only if 𝑝2 − 4𝑞 = 1 . Perhaps there is a typo in the question and the first equality in 𝑝2 = 4𝑞 2 = (1 + 2𝑞)2 is supposed to be replaced by +. Question #7 Prove this has real roots (𝑎2 − 𝑏𝑐)𝑥 2 + (𝑎 + 𝑏)(𝑎 − 𝑐)𝑥 + 𝑎(𝑏 − 𝑐) = 0 Answer: Note that -1 is a root, so the other root is also real. It turns out to be for what you need to prove. 𝑎(𝑏−𝑐) 𝑏𝑐−𝑎2 , but that's not really necessary If you put 𝑥 = −1 , in the equation you would see the LHS equals to 0, it means 0 is one of the root of the equation. Also, a quadratic equation have three possibilities of roots existence. 1. Real and distinct(both) 2. Real and equal 3. Imaginary roots(occurs in conjugate form, means if one is imaginary, other should be also an imaginary root) In our case, we have obtained a real root, so other root must be real. Hence the given equation should have real roots, I think method of discriminant is a bit difficult. You can go by using one root method. Question #8 Prove: if 𝑝𝑥 2 − (𝑝 + 𝑞)𝑥 + 𝑝 = 0 has an integer root, 𝑝𝑥 2 + 𝑞𝑥 + 𝑝2 − 𝑞 = 0 has an integer root too, for com' 𝑝, 𝑞 ∈ 𝑁 Answer: Step 1 In this answer, p and q are co' integers such that 𝑞 ≥ 0 (and p can be zero or negative). We want to show that if 𝑝𝑥 2 − (𝑝 + 𝑞)𝑥 + 𝑝 = 0 has an integer solution, then so does 𝑝𝑥 2 + 𝑞𝑥 + 𝑝2 − 𝑞 = 0 If 𝑥 = 𝑡 is an integer root of 𝑝𝑥 2 − (𝑝 + 𝑞)𝑥 + 𝑝 = 0 , then 𝑡(𝑝 + 𝑞 − 𝑝𝑡) = 𝑝. If 𝑡 = 0 , then 𝑝 = 0 . Because 𝑔𝑐𝑑(𝑝, 𝑞) = 1 , this means 𝑞 = 1 . Then, the equation 𝑝𝑥 2 + 𝑞𝑥 + 𝑝2 − 𝑞 = 0 which is equivalent to 𝑥 − 1 = 0 , has an integer solution 𝑥 = 1 From now on, suppose that 𝑡 ≠ 0 . That is, 𝑝 = 𝑠𝑡 for some integer s. Hence, 𝑝 + 𝑞 − 𝑝𝑡 = 𝑝 𝑡 = 𝑠. Therefore, 𝑞 = 𝑠 + 𝑝(𝑡 − 1) = 𝑠 + 𝑠𝑡(𝑡 − 1) = 𝑠(𝑡 2 − 𝑡 + 1). Therefore, s divides p and q, which implies 𝑠 = ±1 as p and q are co'. Since 𝑞 ≥ 0 , we get 𝑠 = 1 Therefore, 𝑝 = 𝑡 and 𝑞 = 𝑡 2 − 𝑡 + 1 . Hence, 𝑝𝑥 2 + 𝑞𝑥 + 𝑝2 − 𝑞 = 𝑡𝑥 2 + (𝑡 2 − 𝑡 + 1)𝑥 + (𝑡 2 − (𝑡 2 − 𝑡 + 1)) That is, 𝑥 = 1 − 𝑡 is an integer solution to the quation 𝑝𝑥 2 + 𝑞𝑥 + 𝑝2 − 𝑞 = 0 The condition 𝑞 ≥ 0 is necessary. Note that 𝑝 = −2 and 𝑞 = −3 satisfies the condition that 𝑝𝑥 2 − (𝑝 + 𝑞)𝑥 + 𝑝 = 0 has an integer solution, but 𝑝𝑥 2 + 𝑞𝑥 + 𝑝2 − 𝑞 = 0 does not have an integer solution. Question #9 Proving a quadratic equation has no integral roots Show that the quadratic equation 𝑥 2 − 7𝑥 − 14(𝑞 2 + 1) = 0 where q is an integer, has no integral real roots. Answer: To prove that no perfect square 𝑘 2 = 7𝑛 + 6 , we work with 𝑘 = 7𝑚 ± 𝑙; 𝑙 ∈ 0,1,2,3 (7𝑚 + 0)2 = 49𝑚2 = 7(7𝑚2 ) + 0 (7𝑚 ± 1)2 = 49𝑚2 ± 14𝑚 + 1 = 7(7𝑚2 ± 2𝑚) + 1 (7𝑚 ± 2)2 = 49𝑚2 ± 28𝑚 + 4 = 7(7𝑚2 ± 4𝑚) + 4 (7𝑚 ± 3)2 = 49𝑚2 ± 42𝑚 + 9 = 7(7𝑚2 ± 6𝑚 + 1) + 2 None of these are of the form 7𝑛 + 6 , thusly it is not possible Question #10 Proving 3(1 − 𝑎 + 𝑎2 )(1 − 𝑏 + 𝑏 2 )(1 − 𝑐 + 𝑐 2 ) ≥ 1 + 𝑎𝑏𝑐 + 𝑎2 𝑏 2 𝑐 2 My task was to prove the question above over real variables. I thought that this minor inequality should help 3(1 − 𝑎 + 𝑎2 )(1 − 𝑏 + 𝑏 2 ) ≥ 2(1 − 𝑎𝑏 + 𝑎2 𝑏 2 ). which is true. By this inequality, the original inequality is converted to (1 − 𝑎𝑏)2 (1 − 𝑐)2 + (𝑎𝑏 − 𝑐)2 + 𝑎𝑏𝑐 ≥ 0 This proves the inequality for 𝑎𝑏𝑐 ≥ 0 . I want to prove this Inequality for 𝑎𝑏𝑐 ≤ 0 . But I couldn't find a solution for 𝑎𝑏𝑐 < 0 . Any extensions for 𝑎𝑏𝑐 < 0 are thankfully accepted. Answer: Your first step leads to a wrong inequality because it does not save the case of the equality occurring: 𝑎 = 𝑏 =𝑐 = 1. After your first step it's enough to prove that: 2(1 − 𝑎𝑏 + 𝑎2 𝑏 2 )(1 − 𝑐 + 𝑐 2 ) ≥ 1 + 𝑎𝑏𝑐 + 𝑎2 𝑏 2 𝑐 2 , which is wrong for 𝑎 = 𝑏 = 𝑐 = 1 . The Vasc's solution. Since 2(𝑎2 − 𝑎 + 1)(𝑏 2 − 𝑏 + 1) ≥ 𝑎2 𝑏 2 + 1, it's enough to prove that: 3(𝑎2 𝑏 2 + 1)(𝑐 2 − 𝑐 + 1) ≥ 2(𝑎2 𝑏 2 𝑐 2 + 𝑎𝑏𝑐 + 1), which is a quadratic inequality of c.