Questions and Answers #16 Quadratics

Questions and Answers Sheet 16 Quadratics Question #1 Sign of 𝑃(𝛥) with 𝑃(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 and 𝛥 = 𝑏 2 − 4𝑎𝑐 Answer: Completing the square differently, this is non-negative when 𝑎 > 0, Δ ≥ 0,2𝑏 − 1 4𝑎 ≥ 0: 1 b b b2 b b2 c 2 P() =  +  − 2  + 2 + 2  − 2 + a a a 4a a 4a a 2 2 b b 2b b − 4ac = 2 − 2  + 2 +  − 2a 4a a 4a 2 b 1  = ( − ) 2 + (2b − ) 2a 4a a Question #2 How do we know that the quadratic 3𝑦 2 − 𝑦 − 12 has real root? (a) Notice the quadratic cannot be factored into the product of two binomials with integer coefficients. Does this mean that the quadratic does not have any real roots? (b) If the answer to part (a) is "no", then explain how we know that the quadratic does have real roots. (c) Suppose the quadratic has roots 𝑦 = 𝑟 and 𝑦 = 𝑠. Find a quadratic with roots 𝑟 + 2 and 𝑠 + 2. Answer: (a) This does not mean the quadratic has no real roots. A clearer example of this is 𝑦 2 − 2 = (𝑦 + √2)(𝑦 − √2). (b) One way to know is to observe that plugging in 𝑦 = 0 and𝑦 = 3 yields 3 ⋅ 02 − 0 − 12 = −12, 3 ⋅ 32 − 3 − 12 = 12. so somewhere between 0 and 3 the quadratic must equal 0. Another way to know is by computing the discriminant, which is Δ = 𝑏 2 − 4𝑎𝑐 = (−1)2 − 4 ⋅ 3 ⋅ (−12) = 145. The quadratic has a real root because the discriminant is nonnegative. (c) If r and s are roots of 3𝑦 2 − 𝑦 − 12, then it follows that 𝑟 + 2 and 𝑠 + 2 are roots of 3(𝑦 − 2)2 − (𝑦 − 2) − 12 which by a little bit of algebra simplifies to 3𝑦 2 − 13𝑦 + 2. Question #3 Finding the range of 𝑦 = 𝑥 2 +2𝑥+4 2𝑥 2 +4𝑥+9 (and 𝑦 = 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 in general) Answer: The question can be easily solved by this technique: As 𝑦 = 𝑥 2 +2𝑥+4 2𝑥 2 +4𝑥+9 Thus, 2𝑦 = 1 − ⇒ 2𝑦 = 2𝑥 2 +4𝑥+9−1 2𝑥 2 +4𝑥+9 . 1 2(𝑥+1)2 +7 Squares can never be less than zero so the minmum value of the function: 2(𝑥 + 1)2 + 7 would be 7, or 1 1 Maximum value of is . 2 2(𝑥+1) +7 7 3 This tells that minimum value of y will be . 7 3 1 From here you can easily tell the maximum and minimum values: 𝑦 ∈ [ , ) 7 2 Question #4 How to convert a quadratic solution to an unusual format "Solve the quadratic equation 3𝑥 2 + 4𝑥 − 5 giving your answer in the form 𝑎 𝑏±√19 , where a and b are integers." Answer: In general, the quadratic polynomial 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 has roots the first root by −𝑏−√𝑏 2 −4𝑎𝑐 −𝑏−√𝑏 2 −4𝑎𝑐 = 1, you get 𝑏 2 −(𝑏 2 −4𝑎𝑐) −2𝑎𝑏−2𝑎√𝑏 2 −4𝑎𝑐 = −𝑏+√𝑏 2 −4𝑎𝑐 2𝑎 −2𝑐 𝑏+√𝑏 2 −4𝑎𝑐 and −𝑏−√𝑏 2 −4𝑎𝑐 2𝑎 . If you multiply . Doing the same with the other root, you for the desired expressions. Note: I'm assuming you can actually do the division because the roots are non zero, which is what happens in your problem. Question #5 A function is a ratio of quadratic functions and has a vertical asymptote x =4 and just one -intercept, x =1 . It is known f that has a removable discontinuity at x =- 1 and lim 𝑓(𝑥) = 2. Evaluate 𝑥→−1 a) f (0) b) lim 𝑓(𝑥) 𝑥→∞ Answer: Step 1 Given: A function is a ratio of quadratic functions. Vertical asymptote at x =4. x -intercept: x =1 Removable discontinuity at x=-1 and lim 𝑓 (𝑥) = 2 𝑥→−1 Step 2 Explanation: When it says that it is a ratio of quadratics that means that the numerator and the denominator are polynomials of the second degree. This means that they both have two roots. To find vertical asymptotes you set the denominator equal to zero and solve for x. Stating the vertical asymptotes gives you a root of the polynomial in the denominator. The part about it having one x intercept gives you a root of the polynomial in the numerator. The part about the removable discontinuity means that the two polynomials share a common root and the limit at that discontinuity gives us an idea about how to find a constant. In our case the top polynomial (without the constant that we will determine in a little bit) is (x-1)(x+1). the bottom polynomial is (x-4)(x+1). where x=-1 is the common root, x=4 is the vertical asymptote and x=1 is the x intercept. So, we have 𝑓(𝑥) = lim 𝑐(𝑥−1)(𝑥+1) (𝑥−4)(𝑥+1) 𝑐(𝑥−1)(𝑥+1) 𝑥→−1 (𝑥−4)(𝑥+1) = = where c is some constant = lim 𝑐(𝑥−1) 𝑥→−1 (𝑥−4) 𝑐(−1−1) (−1−4) −2𝑐 −5 =2 (Given) So, c=5 and 𝑓(𝑥) = 5(𝑥−1)(𝑥+1) (𝑥−4)(𝑥=1) (a) So, 𝑓(0) = 5 4 (b) The degree of the two polynomials is the same so the limit is 5. Question #6 An inequality constraint problem min 1 2 1 (𝑥1 − 1)2 + (𝑥2 − 2)2 2 𝑠. 𝑡. 𝑥1 − 𝑥2 = 1 and 𝑥1 + 𝑥2 ≤ 2 Answer: Write 𝑥2 = 𝑡 and 𝑥1 = 𝑡 + 1 , then you have to find a minimum of 𝑓(𝑡) = 𝑡 2 + (𝑡 − 2)2 = 2𝑡 2 − 4𝑡 + 4 1 where 𝑡 ≤ . Since it global minimum is at t=1 we see that it is decreasing on (−∞, 1), so 𝑓(𝑡) ≥ 2 1 𝑓( ) = ⋯ 2 Question #7 A basic doubt while taking square roots of variables which have ± sign before the square root sign. Answer: In your section where you come up with 𝛼1 + 𝛽2 = 0, you have: 𝛼1 = sec(𝜃) + |tan(𝜃)| 𝛽1 = sec(𝜃) − |tan(𝜃)| Also we then have 𝛼2 = sec(𝜃) + |tan(𝜃)| and 𝛽2 = − sec(𝜃) +∣ tan(𝜃) So 𝛼1 + 𝛽2 = 2 ∣ tan(𝜃) ∣= −2 tan(𝜃) (last equality because, as you noted, in the fourth quadrant, tangent takes on negative values). Question #8 Proving that 𝑥 2 − 2𝑥𝑦 + 6𝑦 2 − 12𝑥 + 2𝑦 + 41 ≥ 0 where x, 𝑦 ∈ 𝑅 Answer: Let z=x-y then 𝑓(𝑥, 𝑦) = 𝑧 2 + 5𝑦 2 − 12𝑧 − 10𝑦 + 41 = (𝑧 − 6)2 + 5(𝑦 − 1)2 Question #9 Value of an algebraic expression in a quadratic equation Given: 𝑥 2 + √𝑚𝑥 + 𝑛 = 0 Answer: (m + 1)(m − 1) − 2(2n − 1) = m2 − 1 − 4n + 2 = m 2 − 4n + 1 = m2 − m + 1 There's no way to further "solve" this problem. There are infinitely many values for m that satisfy the given conditions. It should be noted that if we are working with a polynomial over the reals, then we must also add the condition that 𝑚 ≥ 0 , since √⋅ cannot input negative values. Question #10 Is an equation only a parabola if it is quadratic? Could the graph of 𝑦 = 𝑥 1.65 be described as parabolic in shape? Answer: Not everything u-shaped is a parabola. A parabola has a very specific shape with particular special properties. For example, for each parabola there is a point called the "focus" and a line called the "directrix" and each point on the parabola is the same distance from the focus as it is from the directrix. This is similar to the way a circle has a center and every point of the circle is an equal distance from the center. A shape might look more or less circular, but if it doesn't have a center that is the same distance from each of its points, it isn't a circle. Curves such as 𝑦 = 𝑥 1.65 , don't have a focus and directrix that behave the way a parabola's do. They might look parabolic, but they are not parabolas. Note also that an equation may have a parabolic graph even if it isn't obvious. The answer below of Vítězslav Štembera says that the equation must have one of two specific forms, but that isn't correct. Any equation of the form 𝐴𝑥 2 + 2𝐵𝑥𝑦 + 𝐶𝑦 2 + 2𝐷𝑥 + 2𝐸𝑦 + 𝐹 = 0 will have a parabolic graph, if it is non-degenerate and if 𝐴𝐶 − 𝐵2 = 0. For example, if one takes the parabola 𝑦 = 𝑥 2 and rotates it by 45∘ the curve is still a parabola, with equation 𝑥 2 − 2𝑥𝑦 + 𝑦 2 − 𝑥√2 − 𝑦√2 = 0.