Questions and Answers #45 Quadratic Function and Equation

Questions and Answers Sheet 45 Quadratic Function and Equation Question #1 How to find the points of intersection in a curve and a line 𝑦 = 8√𝑥 − 2𝑥 Answer: Step 1 Differently from the previous comments, I propose another way to approach it for the sake of curiosity. 8√𝑥 − 2𝑥 = 6 ⟺ 𝑥 − 4√𝑥 + 3 = 0 ⟺ (𝑥 − √𝑥) − (3√𝑥 − 3) = 0 ⟺ √𝑥(√𝑥 − 1) − 3(√𝑥 − 1) = 0 ⟺ (√𝑥 − 3)(√𝑥 − 1) = 0 ⟺ (√𝑥 = 3) ∨ (√𝑥 = 1) ⟺ (𝑥 = 9) ∨ (𝑥 = 1) and we are done. Question #2 How to find the roots in a quadratic function in vertex form? 2( x − ( 2 2 ) −3− 2 2 Answer: Step 1 While you can get it into standard form by using FOIL or the binomial theorem to square (𝑥 − ℎ)2 = 𝑥 2 − 2ℎ𝑥 + ℎ2 it's actually easier not to go through the standard form: 𝑎(𝑥 − ℎ)2 + 𝑘 = 0 𝑎(𝑥 − ℎ)2 = −𝑘 (𝑥 − ℎ)2 = − 𝑘 𝑎 𝑥 − ℎ = ±√− 𝑘 𝑎 𝑥 = ℎ ± √− 𝑘 𝑎 Question #3 How to find the smallest 𝑎, 𝑏 ∈ 𝑁 that solve a single equation 𝑎2 + 𝑎 = 𝑏 2 + 19𝑏 − 12 Answer: Step 1 If 𝑎2 + 𝑢𝑎 = 𝑏 2 + 𝑣𝑏 + 𝑤 Then 4𝑎2 + 4𝑢𝑎 = 4𝑏 2 + 4𝑣𝑏 + 4𝑤 so (2𝑎 + 𝑢)2 − (2𝑏 + 𝑣)2 = 𝑢2 − 𝑣 2 + 4𝑤 so (2𝑎 + 𝑢)2 − (2𝑏 + 𝑣)2 = 𝑢2 − 𝑣 2 + 4𝑤 or 𝑢2 − 𝑣 2 + 4𝑤 = (2𝑎 + 𝑢 + 2𝑏 + 𝑣)(2𝑎 + 𝑢 − 2𝑏 − 𝑣) . All solutions come from factoring 𝑢2 − 𝑣 2 + 4𝑤 = 𝑝𝑞 , and solving 𝑝 = 2𝑎 + 𝑢 + 2𝑏 + 𝑣, 𝑞 = 2𝑎 + 𝑢 − 2𝑏 − 𝑣 Step 2 𝑝 + 𝑞 = 4𝑎 + 2𝑢𝑠𝑜𝑎 = 𝑝+𝑞−2𝑢 𝑝 − 𝑞 = 4𝑏 + 2𝑣 so 𝑏 = 4 and 𝑝−𝑞−2𝑣 4 . Therefore, 𝑢 = 1, 𝑣 = 19, 𝑤 = −12 𝑢2 − 𝑣 2 + 4𝑤 = 1 − 192 − 48 = −408 = −23 ⋅ 3 ⋅ 17 Look at all the factorizations, remembering that p and q can be negative as well as positive and 4 must divide the two expressions. Question #4 How to generate factors given the b-value of the quadratics is the hypotenuse of a Pythagorean Triple? Given two quadratics with integer coefficients, 𝑦 = 𝑥 2 + 𝑏𝑥 + 𝑐 and 𝑥 2 + 𝑏𝑥 − 𝑐 Answer: Step 1 Let p ( x) = x2 + bx  c  [ x] , with discriminant 𝑏 2 ∓ 4𝑐 . As you observed, if 𝑝± have integer roots, then for integers m,n we have 𝑏 2 + 4𝑐 = 𝑚2 , 𝑏 2 − 4𝑐 = 𝑛2 so that 𝑚2 + 𝑛2 = 2𝑏 2 This does not prove that b is the hypotenuse in a pythagorean triple. For example, the degenerate case m=n=b always solves the equation 𝑚2 + 𝑛2 = 2𝑏 2 , but obviously not every positive integer is the hypotenuse in a pythagorean triple. Question #5 How to isolate a in this equation? I have the following equation that I would like to isolate the variable a from, but I am a bit stuck. The variable k is known; however, x and a are unknowns. 𝑎 𝑥 2 + (1−𝑎) 𝑥 − 1 2(1−𝑎)𝑘 =0 Answer: Multiply the first equation by 1 − 𝑎 to get (1 − 𝑎)𝑥 2 + 𝑎𝑥 − 1 2𝑘 = 0. This is linear in a; collecting like terms yields (𝑥 − 𝑥 2 )𝑎 + 𝑥 2 − and hence if 1 −𝑥 2 𝑎 = 2𝑘 𝑥−𝑥 2 = 1 = 0, 2𝑘 x{0,1} then 1−2𝑘𝑥 2 2𝑘𝑥−2𝑘𝑥 2 . Question #6 How to prove 4 cos(𝑥) + 𝑥 sin(𝑥) ≥ 4 − 𝑥 2 for all real x, with equality if and only if 𝑥 = 0 ? Answer: Step 1 Let ℎ(𝑥) = 4 cos(𝑥) + 𝑥 sin(𝑥) + 𝑥 2 − 4 ; note that h(x) is even. For reference, we have ℎ′ (𝑥) = 2𝑥 + 𝑥 cos(𝑥) − 3 sin(𝑥) ℎ′′ (𝑥) = 2 − 2 cos(𝑥) − 𝑥 sin(𝑥) ℎ(3) (𝑥) = sin(𝑥) − 𝑥 cos(𝑥) ℎ(4) (𝑥) = 𝑥 sin(𝑥) ℎ(5) (𝑥) = sin(𝑥) + 𝑥 cos(𝑥) ℎ(6) (𝑥) = 2 cos(𝑥) − 𝑥 sin(𝑥) Claim: It suffices to show ℎ(𝑥) ≥ 0 on [0, 𝜋] We have ℎ′ (𝑥) = 𝑥(1 + cos(𝑥)) + (𝑥 − 3 sin(𝑥)) For 𝑥 > 0 the first term is non-negative and for 𝑥 > 𝜋 the second one iss as well the unique positive real solution of 𝑥 = 3 sin(𝑥) is 𝑥 ≈ 2.28 but we take 𝑥 = 𝜋 for convenience). In other words, for 𝑥 > 𝜋 we have ℎ′ (𝑥) > 0 and ℎ(𝜋) > 0 . Claim: ℎ′ (0) = 0 (obvious) and h has its global minimum at zero. This is basically a corollary to the Second Derivative Test. Note that ℎ′′′′ (𝑥) ≥ 0 on [−𝜋, 𝜋] as the signs of the terms agree, which implies h'' has a minimum at 𝑥 = 0 , i.e. that ℎ′′ (𝑥) ≥ 0 on [−𝜋, 𝜋] . Alternatively, since the sixth derivative is the first to evaluate to a nonzero number at zero and this value is positive, h has a minimum at 𝑥 = 0 . Question #7 How to prove that if a quartic equation ( with real coefficients ) has 4 imaginary roots they all will be in conjugate pairs? I proved this fact for qudratic equation in the following way , let a qudratic equation have a imaginary root 𝑝 + 𝑖𝑞 (q is not 0), and let other root be (𝑎 + 𝑖𝑏). Now here sum of roots will be a real number lets say R, ⇒ (𝑝 + 𝑖𝑞)(𝑎 − 𝑖𝑞) = 𝑅′ ⇒ (𝑝𝑎 + 𝑞 2 ) + 𝑖(𝑎𝑞 − 𝑝𝑞) = 0 ⇒ 𝑎𝑞 − 𝑝𝑞 = 0 ⇒ 𝑎𝑞 = 𝑝𝑞 ⇒ 𝑎 = 𝑝 . So finally the roots are, 𝑝 + 𝑖𝑞 and 𝑝 − 𝑖𝑞 , hence proved. I tried to prove this fact for quartic equation in a similar way but could not reach to the conclusion. Please guide me by answering by my method or by suggesting any other simple method to prove that if a quartic equation has all imaginary roots then they will occur in conjugate pairs. Answer: If 𝑃(𝑧) = ∑𝑛𝑖=0 𝑎𝑖 𝑧 𝑖 is a polynomial having real coefficients with 𝑧0 as a root, then taking conjugate on both sides we see that 𝑧0 is also a root. Now, if 𝑎 + 𝑖𝑏 is a root of a quartic with real coefficients, then by this reasoning, 𝑎 − 𝑖𝑏 is also a root. Hence, given polynomial is divisible by the quadratic (𝑥 − 𝑎 − 𝑏𝑖)(𝑥 − 𝑎 + 𝑏𝑖) = (𝑥 − 𝑎)2 + 𝑏 2 . Divide it by this quadratic, get a quadratic and apply the same procedure. Question #8 How to show if a linear function through a point passed through a curve twice? Show that any linear function passing through 𝑃(0,3) will meet the curve 𝑓(𝑥) = 2𝑥 2 − 𝑥 − 2 twice Answer: The discriminant is: 𝐷 = 𝑏 2 − 4𝑎𝑐 Note that this value doesn't tell you the roots but rather tells you how many there are. If 𝐷 > 0 , then there are 2 roots. If 𝐷 = 0 , then there is 1 root. If 𝐷 < 0 , then there are no roots. Question #9 How to solve 𝑒𝐵 2 𝜔4 − [( ) + 2𝜔02 ] 𝜔2 + 𝜔04 = 0 𝑚 in the simplest way Answer: Step 1 2 2 2 𝑒𝐵 𝑒𝐵 4𝜔2 = 2 ( ) + 4𝜔02 ± 2√[( ) + 2𝜔02 ] − 4𝜔04 𝑚 𝑚 2 4 2 𝑒𝐵 𝑒𝐵 𝑒𝐵 = 2 ( ) + 4𝜔02 ± 2√( ) + 4 ( ) 𝜔02 𝑚 = 2( =[ 𝑚 𝑒𝐵 2 𝑒𝐵 𝑒𝐵 2 ) + 4𝜔02 ± 2 ( ) √( ) + 4𝜔02 𝑚 𝑚 𝑚 𝑒𝐵 𝑚 𝑚 𝑒𝐵 2 ± √( ) + 𝑚 2 4𝜔02 ] so it gives you the "nice" formula for the roots of 𝜔2 ± 𝑒𝐵𝜔 𝑚 − 𝜔02 Question #10 How to solve a quadratic equation with Binary coefficients? 5𝑥 2 − 50𝑥 + 125 = 0 It has the roots 𝑥1 = 5 and 𝑥2 = 5 . But now, convert these coefficients into binary: 101𝑥 2 − 110010𝑥 + 1111101 = 0 Solve this quadratic equation with binary coefficients. Answer: Just because the coefficients of the quadratic equation are in a different base doesn't mean that its roots are different! In decimal, the roots are: 𝑥1 = 5 𝑥2 = 5 and because 5 = (101)2 in binary, thus the roots of the equation (101)2 𝑥 2 − (110010)2 𝑥 + (1111101)2 = 0 are: 𝑥1 = (101)2 𝑥2 = (101)2