Questions and Answers #69 Quadratic Function and Equation

Questions and Answers Sheet 69 Quadratic Function and Equation Question #1 Solving for y, given 𝑥= −1+√1+(4𝑦/50) 2 Answer: This is the formula for the quadratic function 𝑓 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐. In this case we have 𝑎 = 1 and 𝑏 = 1 . Comparing 4𝑎𝑐 = 4𝑐 with − Thus 𝑓 = 𝑥 2 + 𝑥 − 𝑦 50 =0 Question #2 Solving 𝑥 2 +12𝑥+4 𝑥+2 = 6√ 𝑥 Answer: Middle term splitting will always remain my favorite method for factorising polynomials! x 2 + 12 x + 4 =6 x x+2 ( x + 2) 2 + 8 x =6 x x+2 ( x + 2) 2 − 6 x ( x + 2) + 8 x = 0    ( x + 2) 2 − 4 x ( x + 2) − 2 x ( x + 2) + 8 x = 0  ( x + 2 − 4 x )( x + 2 − 2 x ) = 0 Now apply the quadratic formula. Question #3 Solving 1 𝑥−2 + 1 𝑥−1 > 1 𝑥 Answer: Actually, 1 𝑥−2 + 1 𝑥−1 1 𝑥 2 −2 𝑥 𝑥(𝑥−1)(𝑥−2) − = . Besides x^2-2>0 if and only if 𝑥 ∈ (−∞, −√2) ∪ (√2, ∞) 𝑥 2 − 2 < 0 if and only if 𝑥 ∈ (−√2, √2) 𝑥(𝑥 − 1)(𝑥 − 2) > 0 if and only if 𝑥 ∈ (0,1) ∪ (2, ∞) 4𝑦 50 𝑦 gives 𝑐 = − . 50 𝑥(𝑥 − 1)(𝑥 − 2) < 0 if and only if 𝑥 ∈ (−∞, 0) ∪ (1,2) Question #4 Solving 2 + 1 cos2 𝑥 + 5 cos 𝑥 =0 Answer: Hint: by multiplying both sides by cos 2 𝑥, this is the same equation as 2 cos 2 𝑥 + 1 + 5 cos 𝑥 = 0, which is a quadratic equation in cos 𝑥. Question #5 Solving 3𝑥 2 + 4𝑥 − 2 = 0( 𝑚𝑜𝑑 31) Answer: Might be easier to factor or use the quadratic formula. 3𝑥 2 + 4𝑥 − 2 ≡ 0 𝑚𝑜𝑑 31 so abusing notation where √𝑘 will mean the congruence a where 𝑎2 ≡ 𝑘 𝑚𝑜𝑑 31 and congruence where 𝑚(𝑚−1 ) ≡ 1 𝑚𝑜𝑑 31 then 𝑥≡ −4±√16+24 6 1 𝑚 = 𝑚−1 is the ≡ (−4 ± √40) ⋅ 6−1 ≡ (27 ± √9) ⋅ 6−1 ≡ (27 ± 3) ⋅ 6−1 ≡ 30  6−1  5  6  6−1  5  −1 −1 24  6  4  6  6  4 So x4 ( mod 31) or ( mod 31) x5 ( mod 31) Question #6 Solving 𝑥 2 − ⌊𝑥⌋ − 4 = 0 Answer: If |𝑥| ≥ 3, then 𝑥 2 > 𝑥 + 4 ≥ ⌊𝑥⌋ + 4, and no solution is possible. Therefore the only solutions have ⌊𝑥⌋ = −3, −2, −1,0,1, and 2. Let's see if ⌊𝑥⌋ = −3 works: 𝑥 2 − (−3) − 4 = 0 𝑥 = ±1 But neither ⌊1 ⌋ nor ⌊−1⌋ equals -3 so we have no solution here Now try ⌊𝑥⌋ = −2 𝑥 2 − (−2) − 4 = 0 𝑥 = ±√2 We see that ⌊√2⌋ ≠ −2 so that's not a solution, but ⌊−√2⌋ = −2 so we have a solution. Check the other 4 cases similarly and find that 𝑥 = √6 is the other solution. Question #7 Solving quadratic equation in motion equations Is it possible to solve this equation for t 1 𝑥𝑓 = 𝑥0 + 𝑣0 𝑡 + 𝑎𝑡 2 ~? 2 It feels like its a quadratic but since x final and initial are not the same and sometimes initial is not zero any idea? Answer: The solution given on physics SE is most likely formatted incorrectly. It should be 𝑡= −𝑣0 ±√𝑣02 −2𝑎(𝑥0 −𝑥𝑓 ) 𝑎 This gives you the solutions t=0, t=1.65 and you can safely ignore the root t=0 . There is a simpler way to solve this equation. If 𝑥0 and 𝑥𝑓 are 0, you can remove them from the equation directly. The equation then becomes 2𝑣0 𝑡 = −𝑎𝑡 2 Ignoring the trivial solution t=0 , 𝑡=− 2𝑣0 𝑎 = 16.2 9.81 ≈ 1.65 In physical terms, your equation of motion describes a ball thrown up with velocity 𝑣0 . You are simply finding the time of flight of the particle in this case. Question #8 Solving quadratic inequalities algebraically without graphing/common-sensing the equality signs I've just been presented an inequality at GCSE (high school) level that, solving algebraically (the method used for linear inequalities), doesn't work correctly which has caught me by surprise. Research always leads to graphed or common-sensed solutions. Can anyone explain why this doesn't work and if there's an algebraic manipulation that leads to the right answers? 𝑥 2 − 49 > 0 Using the quadratic identity (x + 7)(x - 7) > 0 Divide both sides by (x+7) ... x-7>0 x>7 Divide both sides by (x-7) ... x+7>0 x>-7 Alternatively, 𝑥 2 − 49 > 0 𝑥 2 > 49 𝑥 > ±7 The negative solution has to be x < -7 but I don't know how that's arrived at through algebra. Answer: It is much fun and pretty enlightening to do this kind of exercises with a ping-pong between algebra and geometry. Anyway, if you want to avoid that , then we have the following algebraic facts which, funny enough, are pretty easy to prove precisely by means of geometry...but not only, as they follow from the definition of absolute value: For For 0  a  | x | a  −a  x  a  ,  | x | a  x  −a or x  a  Thus, in your case, we could do as follows (all algebra, not geometry): x2 − 49  0  x2  49 taking square roots  | x | 7  x  −7 or x  7 and in intervals notation, the solution is 𝑥 ∈ (−∞, −7) ∪ (7, ∞) . Question #9 Solving quadratic inequality 𝑥 2 > 0 Answer: Step 1 Caution, 𝑎2 > 𝑏 does not imply 𝑎 > ±√𝑏 But ±𝑎 > √𝑏 is correct. (With a somewhat sloppy notation.) Step 2 More rigorously 𝑎2 > 𝑏 ⟺ (𝑎 − √𝑏)(𝑎 + √𝑏) > 0 ⟺ (𝑎 > √𝑏 ∧ 𝑎 ≻ √𝑏) ∨ (𝑎 < √𝑏 ∧ 𝑎 < −√𝑏) ⟺ 𝑎 > √𝑏 ∨ 𝑎 < −√𝑏. Question #10 Solving simultaneous equations with multiple variables. If (𝑥 + 𝑎)2 is a factor of 𝑥 3 + 6𝑝𝑥 + 𝑘, show that 𝑘 + 2𝑎3 = 0 I've tried different ways to solve this, but is it even possible? Answer: Let p(x) denote the 3rd degree polynomial. Your condition means that -a is a double root of p, therefore it is also a root of the derivative 𝑝(𝑥) = 3𝑥 2 + 6𝑝. As a consequence, we have:  p(a) = −a 3 − 6 pa + k  2  p(a) = 3a + 6 p = 0 from which you easily derive your constraint. = 0