Questions and Answers #39 Quadratic Function and Equation

Questions and Answers Sheet 39 Quadratic Function and Equation Question #1 2 1 Finding the roots of 𝑥 3 + 𝑥 3 − 2 Answer: 2 1 I presume the equation is 𝑥 3 𝑥 3 − 2 = 0 How have you reached to 𝑥 2 + 𝑥 − 8 = 0 Better use the following substitution: 1 𝑥 3 = 𝑦 ⇒ 𝑥 = 𝑦3 1 2 2 𝑥 3 = (𝑥 3 ) =? to form a quadratic equation y Question #2 Finding the value of a variable in a quadratic The number of negative integral values of m for which the expression 𝑥 2 + 2(𝑚 − 1)𝑥 + 𝑚 + 5 is positive ∀𝑥 > 1 is? Answer: There are two ways for having p( x) = x 2 + 2(m − 1) x + m + 5  0 x  1 Either p(x) has no real root, which means the reduced discriminant Δ′ = (𝑚 − 1)2 − (𝑚 + 5) = (𝑚 + 1)(𝑚 − 4) < 0 is negatve; or p(x) has real roots, say 𝜉0 ≤ 𝜉1 , 𝑖. 𝑒. Δ′ ≥ 0 In this case, the condition means 1 is on the right of the interval of the roots. This is satisfied if and only if 𝑝(1) ≥ 0 (1 is outside of the interval of the roots or is a root) and, using Vieta's relations, 1 1 > (𝜉0 + 𝜉1 ) = 1 − 𝑚 2 Question #3 Finding the vertex of a quadratic function: 5𝑥 2 − 20𝑥 + 15 Answer: Let's say you have a quadratic function 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 Then, you create a new quadratic function by multiplying by a factor of \lambda : 𝑔(𝑥) = 𝜆𝑓(𝑥) = 𝜆𝑎𝑥 2 + 𝜆𝑏𝑥 + 𝜆𝑐 Putting f into vertex form, we have: 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 𝑏 𝑦 = 𝑎 (𝑥 2 + 𝑥) + 𝑐 𝑎 𝑏 𝑏2 𝑎 4𝑎2 𝑦 = 𝑎 (𝑥 2 + 𝑥 + 𝑦 = 𝑎 (𝑥 − (− 𝑦 − (𝑐 − 𝑏2 4𝑎 2 𝑏 2𝑎 𝑏2 )− )) − 4𝑎 𝑏2 4𝑎 +𝑐 +𝑐 ) = 𝑎 (𝑥 − (− 2 𝑏 2𝑎 )) Comparing this to vertex form, 𝑦 − 𝑘 = 𝑎(ℎ − 𝑘)2 , we have the following coordinates for the vertex of f: (ℎ𝑓 , 𝑘𝑓 ) = (− 𝑏 2𝑎 ,𝑐 − 𝑏2 4𝑎 ). Now, let's consider what happens to the vertex if we scale our quadratic f by a factor of 𝜆: 𝑔(𝑥) = 𝜆𝑓(𝑥) = 𝜆𝑎𝑥 2 + 𝜆𝑏𝑥 + 𝜆𝑐. Following the same process of completing the square, we have: 𝑦 = 𝜆𝑎𝑥 2 + 𝜆𝑏𝑥 + 𝜆𝑐 𝑏 𝑦 = 𝜆𝑎 (𝑥 2 + 𝑥) + 𝜆𝑐 𝑎 𝑏 𝑏2 𝑎 4𝑎2 𝑦 = 𝜆𝑎 (𝑥 2 + 𝑥 + 𝑦 = 𝜆𝑎 (𝑥 − (− 𝑦 − 𝜆 (𝑐 − 𝑏2 4𝑎 𝑏 2𝑎 )− 2 )) − 𝜆𝑏 2 4𝑎 𝜆𝑏 2 4𝑎 + 𝜆𝑐 ) = 𝜆𝑎 (𝑥 − (− 𝑦 − 𝜆ℎ𝑓 = 𝜆𝑎(𝑥 − 𝑘𝑓 ) + 𝜆𝑐 𝑏 2𝑎 2 )) 2 Notice, the vertex of g is at coordinates (ℎ𝑔 , 𝑘𝑔 ) = (𝜆ℎ𝑓 , 𝑘𝑓 ). So, scaling our function by a factor of 𝜆 leaves the x-coordinate of the vertex unchanged, but it does scale the y-coordinate of the vertex by that same factor. 1 In this case, that scaling factor was 𝜆 = , so the x-coordinate was the same for both: 5 ℎ𝑓 = ℎ𝑔 = 2. But, the y-coordinate of the scaled function was 1 1 5 5 1 5 the y-coordinate of the original function: ℎ𝑔 = −1 = × −5 = × ℎ𝑓 . Question #4 Finding the vertex of 𝑦 = 𝑥 2 − 6𝑥 + 1. My solution doesn't match the book. Answer: Any general parabola can be written as 𝐿21 = 4𝐴𝐿2 , if 𝐿1 and 𝐿2 are Eq. of non-parallel lines. If 𝐿1 and 𝐿2 are perpendicular and 𝐿1 and 𝐿2 are normalized, life is simple. The length of latus rectum (LR) is 4A. The Eq. of axis of the parabola is 𝐿1 = 0 Tangen at vertex is 𝐿1 = 0. Vertex is found by solving 𝐿1 = 0 and 𝐿2 = 0. The Eq. of directrix is 𝐿2 = −𝐴, Eq. of LR is 𝐿2 = 𝐴, the focus F come by solving 𝐿1 = 0 & 𝐿2 = 𝐴. 1 So in your case 𝑦 = 𝑥2 − 6𝑥 + 1 ⇒ 𝑦 + 8 = (𝑥 − 3)2. Then 𝐿1 = 𝑥 − 3, 𝐿2 = 𝑦 + 8, 𝐴 = . The vertex is 4 given by 𝐿1 = 0, 𝐿2 = 0 which is x=3, y=-8. The vertex is (3,-8). Question #5 Finding the x values of a quadratic that has many different variables The two solutions of the quadratic equation 3𝑥 2 − 𝑥 + 𝑘 = 0 are 𝑝 4 and 𝑝 + 1. Determine the values of k and p. Answer: We know the two solutions: 𝑝 4 and p+1 So, just substitute x in the equation with the above values. 𝑝 2 𝑝 4 4 3( ) − +𝑘 = 0 and, 3(𝑝 + 1)2 − (𝑝 + 1) + 𝑘 = 0 Question #6 For how many integer values of m parabola 𝑦 = (𝑚 − 2)𝑥 2 + 12𝑥 + 𝑚 + 3 passes through only three quadrants? 1) 0 2) 7 3) 12 4) infinity Answer: One mistake is that 𝑚2 + 𝑚 − 42 < 0 gives 𝑚 ∈ (−7,6) instead of 𝑚 ∈ [−7,6]. So {−7, 6} are excluded. If one excludes also m=2 (which does not give a parabola), then there are only 7 possible choices: {−6, −5, −4, −3, 3, 4, 5} Question #7 For positive integers x, y, and positive real k, if 𝑥+1 𝑦+𝑘 𝑦 = , then 𝑘 ≥ 1. 𝑥 Answer: 𝑎2 +𝑎−𝑏 2 𝑏 ⇒ =𝑐 𝑎2 +𝑎−𝑏 2 𝑏 >0 ⇒ 𝑎2 + 𝑎 − 𝑏 2 > 0 ⇒ 𝑎2 + 𝑎 > 𝑏 2 ⇒ (𝑎 + 1)2 > 𝑏 2 ⇒ 𝑎+1 >𝑏 ⇒ 𝑎 ≥ 𝑏 (for integers x, y if x+1>y then 𝑥 ≥ 𝑦 ) ⇒𝑐= 𝑎2 +𝑎−𝑏 2 𝑏 ≥ 𝑏 2 +𝑏−𝑏 2 𝑏 =1 Question #8 For which a, b, c is the diophantine quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 𝑦 2 double? Answer: Solve for x to get 𝑥 = −𝑏±√𝑏 2 −4𝑎𝑐+4𝑎𝑦 2 2𝑎 so you need 𝑏 2 − 4𝑎𝑐 + 4𝑎𝑦 2 to be a square, lets say 𝑏 2 − 4𝑎𝑐 + 4𝑎𝑦 2 = 𝑧 2 If a is negative it's easy to see that you need only a finite search. That's also the case if a is a square. The remaining cases are studied under the name of generalized Pell's equation Question #9 For which y is 3 𝑥 4 − 4𝑥 2 𝑦 + 𝑦 2 < 0 How does one show that the function 𝑓(𝑥, 𝑦) = 3𝑥 4 − 4𝑥 2 𝑦 + 𝑦 2 is negative for 𝑥 2 < 𝑦 < 3𝑥 2 Answer: (𝑦 − 2𝑥 2 )2 − 𝑥 4 < 0 (𝑦 − 3𝑥 2 )(𝑦 − 𝑥 2 ) < 0 𝑥 2 < 𝑦 < 3𝑥 2 , which is possible for any 𝑥 ≠ 0 . Question #10 From 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 (quadratic formula) 3 points (5,500), (3,375), (12,980) I solved 25a+5b+c=500 9a+3b+c=375 144a+12b+c=980 s it possible to solve it without a calculator (for the simultaneous equations) If yes, How can I do it? Answer: 1) 25a+5b+c=500 2) 9a+3b+c=375 3) 144a+12b+c=980 4) (1) − (2) → 16𝑎 + 2𝑏 = 125 5) (3) − (1) → 119𝑎 + 7𝑏 = 480 6) 7 × (4) → 112𝑎 + 14𝑏 = 875 7) 2 × (5) → 238𝑎 + 14𝑏 = 960 (7)-(6) → 126𝑎 = 85 → 𝑎 = 85 126 Go back to (4) to get b and then to (1) to get c.