Questions and Answers #17 Quadratics

Questions and Answers Sheet 17 Quadratics Question #1 Are there any two real nonzero numbers a and b such that 𝑎2 + 𝑎𝑏 + 𝑏 2 = 0. Why or why not? Answer: Fix 𝑏 ≠ 0 and try to solve in a. This a second order equations so the possible values of a in terms of b are 𝑎1,2 = −𝑏∓√−3𝑏 2 2 So, for any non zero real value b the possible values of a are complex conjugated because −3𝑏 2 < 0 Therefore you cannot find two non-zero reals such that (𝑎 + 𝑏)2 = 𝑎𝑏 Question #2 Find the value of the product of roots of this quadratic equation It is given that one of the roots of the quadratic equation : 𝑥 2 + (𝑝 + 3)𝑥 − 𝑝2 = 0 Answer: Step 1 Without applying the factorization of a "difference of two squares" or Viete's relations, we can still use the information stated in the problem. If we call the two roots of the quadratic equation r and -r , then we have 𝑟 2 + (𝑝 + 3) × 𝑟 − 𝑝2 = 0 and [−𝑟]2 + (𝑝 + 3) × [−𝑟] − 𝑝2 = 𝑟 2 − (𝑝 + 3) × 𝑟 − 𝑝2 = 0 This means that 𝑟 2 = −(𝑝 + 3) × 𝑟 + 𝑝2 = (𝑝 + 3) × 𝑟 + 𝑝2 ⇒ 2 × (𝑝 + 3) × 𝑟 = 0 So either r=0 or p=-3 But if r=0=-r, then 02 + (𝑝 + 3) × 0 − 𝑝2 = 0 would require p=0, which would then make the quadratic equation 𝑥3 + 3 × 𝑥 = 0 But that polynomial factors as 𝑥 × (𝑥 + 3) = 0, so we couldn't have both roots equal to zero. Instead, it must be that p=-3 , making the equation 𝑥 2 + 0 × 𝑥 − (−3)2 = 𝑥 2 − 9 = 0, for which the roots are given by 𝑟 2 = 9 ⇒ 𝑟 = +3, − 3; the product of the roots is thus -9 Step 2 Another way to arrive at this conclusion is that 𝑦 = 𝑥 2 + (𝑝 + 3) × 𝑥 − 𝑝2 is the equation of an "upward-opening" parabola, for which we want the x-intercepts to be x=-r and x=r . Its axis of symmetry is located midway between these intercepts, so we have h=0 in the "vertex form" of the parabola's equation, 𝑦 = (𝑥 − 0)2 − 𝑝2 (The vertex is definitely "below" the x-axis at (0, − 𝑝2 ), so we know these x-intercepts exist.) The equation of the parabola is therefore 𝑦 = 𝑥 2 − 𝑝2 , making p+3=0 and the rest of the argument above follows. Question #3 Factoring 2𝑘 2 + 7𝑘 + 6 Answer: An easy way to solve the factoring problem is to recast it as a system of equations. Because 2 is a ' number, you know that it's going to look like this: 2𝑘 2 + 7𝑘 + 6 = (2𝑘 + 𝑎)(𝑘 + 𝑏) where a, b are unknown. Expand to get: 2𝑘 2 + 7𝑘 + 6 = 2𝑘 2 + (𝑎 + 2𝑏)𝑘 + 𝑎𝑏 Thus, 𝑎 + 2𝑏 = 7 , ab=6 Now, play with factors of 6 (1,2,3,6). Using a bit of trial and error should get you the answer. Question #4 Integer between the 's p and p+2 is a perfect square. Explain why there exists an integer n such that 4𝑛2 − 1 = 𝑝 a) Is the integer between p and p+2 odd or even? b) Assume additionally that the integer between the 's p and p+2 is a perfect square. c) By considering (2n-1)(2n+1) , find the only possible value of p.v Answer: a) If p=2 then p+2=4 is not ', i.e. we can be sure that the two 's p and p+2 are odd. b) If 𝑝 + 1 = 𝑛2 with positive integer n, then 𝑝 = 𝑛2 − 1 = (𝑛 + 1)(𝑛 − 1) is a factorisation of p. Conclude that n-1=1 i.e., p=3 It turns out that in the only possible case, the number between 3 an 5 is also of the form 4𝑛2 Question #5 Find real solution to the equation 3𝑥 2 + 3𝑦 2 − 4𝑥𝑦 + 10𝑥 − 10𝑦 + 10 = 0. What does partial differentiation give for a second degree equation which doesn't represent a conic? Answer: A conic curve, is the intersection of the 3D cone 𝑝1 𝑧 2 = 𝑝2 𝑥 2 + 𝑝3 𝑦 2 with the 3D plane 𝑝4 𝑥 + 𝑝5 𝑦 + 𝑝6 = 0 which takes the general form as stated in the OP. Now, a point can also be regarded as a conic curve (or conic, simply) in a sense that it is the intersection of the 3D plane with the vertex of the 3D cone, i.e. the point in which, the two halves of the cone meet. An example is the intersection of 𝑧 2 = 𝑥 2 + 𝑦 2 with 2𝑧 = 𝑥 + 𝑦 Hence, your approach is correct. Another way for justifying the result, is to note that any single point is an ellipse with zero diameters and ellipse is a conic curve. Question #6 Solving quadratic matrix equation for X 1 a  X2 =   0 1  𝑅 where 𝑎 ∈ . Solve for matrix X. 0 Answer: Let 1 a  M (a) =  . 0 1  Note M(a) has one eigenvalue: 1, and is not diagonalisable. This means that, if 𝑋 2 = 𝑀(𝑎), then X has eigenvalue 1 or -1, but not both (if it had both, then X would be diagonalisable, and so would 𝑋 2 = 𝑀(𝑎)) We also note that any eigenvectors of X will be an eigenvector for 𝑋 2 , and 𝑋 2 has only the eigenvector 1  0   Thus, X must have only this eigenvector as well. (Of course, when I say a matrix has only one eigenvector, I mean that there is one eigenspace, and it is one-dimensional, so we can only find one linearly independent eigenvector.) 1  0 Now, multiplication by   on the right reveals the first column of a 2 × 2 matrix. Therefore, the first column of X is 1  1  1 X   = 1   =   , 0  0   0  depending on whether the unique eigenvalue is 1 or -1. This makes the matrix X upper-triangular (due to the 0 in the bottom left corner), so the eigenvalues of X lie on the diagonal. That is, X takes the form: 1 *  1 *  0 1 =  0 1 ,     where the three ±′ 𝑠 are all the same. That is, 𝑋 = ±𝑀(𝑏) 2 for some b. Note that (±𝑀(𝑏)) = 𝑀(𝑏)2 , so we are looking for b such that 2 1 b  1 a  0 1  = 0 1  ,     𝑎 which was solved by Angel in their answer: 𝑏 = . This gives us precisely two solutions: 2  1 X =  0 a  1 2,−   1  0 a 2.  1 Question #7 Why do I get incorrect solutions x=0 and x=2 for 𝑥 + 1 + √4𝑥 + 1 = 0? Answer: Here is one correct way to solve this problem I learnt from comments. I should write 𝑥 + 1 = −√4𝑥 + 1 ( x + 1) 2 = 4 x + 1  ( x + 1)  0  x( x − 2) = 0   x  −1 Then I should solve x(x-2)=0 to get x=0 and x=2 . Then I should reject both solutions because both solutions do not satisfy 𝑥 ≤ 1 . Therefore the equation has no solutions. Question #8 Determine the real number A such that the following equation has a real answer for C for every real 𝐵: 𝐴2 + 𝐵2 + 𝐶 2 + 2𝐴𝐵𝐶 = 1 Answer: Solve for C: 𝐶 = −𝐴𝐵 ± √𝐴2 𝐵2 − 𝐴2 − 𝐵2 + 1 We need a real A such that 𝐴2 𝐵2 − 𝐴2 − 𝐵2 + 1 ≥ 0 for any real B. Notice that we can factor this to get (𝐴2 − 1)(𝐵2 − 1) ≥ 0. Since 𝐵2 − 1 ≥ 0 for all |𝐵| ≥ 1 and 𝐵2 − 1 < 0 for all |B|<1 there are only two A-values that can maintain the above inequality regardless of the B-value; that is, 𝐴 = ±1 Question #9 Let p, q, and r be constants. One solution to the equation (𝑥 − 𝑝)(𝑥 − 𝑞) = (𝑟 − 𝑝)(𝑟 − 𝑞) is x=r. Find the other solution in terms of p, q, and r. Answer: You can use Vieta's Theorem. Your equation is equal to 𝑥 2 − (𝑝 + 𝑞)𝑥 − 𝑟(𝑟 − (𝑝 + 𝑞)) = 0 One of the solutions is r hence one root of this polynomial is r. Let y be the second root, we get: r+y=p+q ry=-r(r-(p+q)) Now set y=-r+p+q and we get (y-p)(y-q)=(-r+q)(-r+p)=(-1)(r-q)(-1)(r-p)=(r-q)(r-p) This shows y is a solution to your equation. Question #10 Find all values of a and b that will make (𝑎𝑥 + 5)2 = 16𝑥2 + 𝑏𝑥 + 25 true for all x. Answer: The quadratic formula is not needed here (𝑎𝑥 + 5)2 = 16𝑥 2 + 𝑏𝑥 + 25 ⇒ 𝑏 = 𝑎2 𝑥 − 16𝑥 + 10𝑎 ⇒ (𝑎𝑥 + 5)2 = 16𝑥 2 + (𝑎2 𝑥 − 16𝑥 + 10𝑎)𝑥 + 25 = (𝑎𝑥 + 5)2 We can see from (𝑎𝑥 + 5)2 = (𝑎𝑥 + 5)2 how all values of x require that 𝑏 = 𝑎2 𝑥 − 16𝑥 + 10𝑎 On the other hand, expanding both sides and comparing coefficients, 𝑎2 𝑥 2 + 10𝑎𝑥 + 25 = 16𝑥 2 + 𝑏𝑥 + 25 these results are easily verified in a spreadsheet. This is also confirmed by 𝑏 = 𝑎2 𝑥 − 16𝑥 + 10𝑎 = 42 𝑥 − 16𝑥 + 10 × 4 =40 Question #11 Solve fractional inequalities that have quadratics 2𝑥 2 +3𝑥−2 2𝑥 2 −3𝑥−2 >0 How do you find the range of solutions? Can you just bring the denominator onto the right hand side or does that lose solutions? Answer: Another way of doing this kind of question is to multiply both sides by the denominator squared which, being non-negative, preserves the inequality. So you could go straight to (2𝑥 2 + 3𝑥 − 2)(2𝑥 2 − 3𝑥 − 2) > 0 ⇒ (2𝑥 − 1)(𝑥 + 2)(2𝑥 + 1)(𝑥 − 2) > 0 You then just have to sketch a quick quartic graph showing the four roots and read off the solution sets. Question #12 Factoring 4𝑥 2 − 2𝑥𝑦 − 4𝑥 + 3𝑦 − 3 Why isn't it working? Answer: When 2x=3 , then the expression becomes 4𝑥 2 − 2𝑥𝑦 − 4𝑥 + 3𝑦 − 3 = 9 − 3𝑦 − 6 + 3𝑦 − 3 = 0 Hence the expression can be factored by (2x-3) , and you can see easily that 4𝑥 2 − 2𝑥𝑦 − 4𝑥 + 3𝑦 − 3 = (2𝑥 − 3)(2𝑥 − 𝑦 + 1) Question #13 The area of a rectangular garden is given by the trinomial 𝑥 2 + 6𝑥 − 27. What are the possible dimensions of the rectangle? Answer: It's important to confirm this, but I believe that finding values for x is as easy as solving the quadratic formula. If so, then all you do is factorise and solve, factorising a quadratic involves two rules the numbers in the brackets have to sum to make b and multiply to make c of the formula 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 9 and -3 multiply to make -27 and add to make 6 therefore (x-3)(x+6) since this is a quadratic and equals zero that tells you that either one of the brackets are equal to zero (as anything multiplied by zero is zero) x-3=0 therefore x=3 x+6 = 0 therefore x=-6