University:
Massachusetts Institute of Technology
Course:
18.06 | Linear Algebra
Academic year:

2024
Views:

436

Pages:

3
Author:

Mckenna Welch

Question 1 How do you solve by factoring and using the principle of zero products x² + 7x + 6 = 0? Answer x = -6 or x = -1 Explanation: using the a-c method the factors of +6 which sum to +7 are +6 and +1 => (x + 6)(x + 1) = 0 equate each factor to zero and solve for x x + 6 = 0 => x = -6 x + 1 = 0 => x = -1 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 : x² + (p + 3)x - p² = 0 Answer 1 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 r² + (p + 3) x r - p² = 0 and [-r]² + (p + 3) x [-r] - p² = r² - (p + 3) x r - p² = 0 This means that r² = -(p + 3) x r + p² = (p + 3) x r + p² = 2 x (p + 3) x r = 0 So either r = 0 or p = -3 But if r = 0 = -r, then 0² + (p + 3) x 0 - p² = 0 would require p = 0, which would then make the quadratic equation x³ + 3 x x = 0 But that polynomial factors as x x (x + 3) = 0, so we couldn't have both roots equal to zero. Instead, it must be that p = -3, making the equation x² + 0 x x - (-3)² = x² - 9 = 0, for which the roots are given by r² = 9 => r = +3, -3; the product of the roots is thus -9 Step 2 Another way to arrive at this conclusion is that y = x² + (p + 3) x 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 = x² - p², making p + 3 = 0 and the rest of the argument above follows. Answer 2 Step 1 We know that this equation has at most two roots in the set of reals. Let's denote them with a and -a. Then your equation is x² - a² = 0. Therefore p must be equal to -3. Hence the equation is x² - 9 = 0 and the roots are 3 and -3. So the desired product is -9. Question 3 How do you write an expression to represent the total volume of two different sized boxes as a sum of cubes if one of the boxes has sides with a length of 1 foot and the others has sides with a length of x feet? Answer Step 1 Volume of 1 ft cube box: V₁ = 1³ Volume of x ft. cube box: V₂ = x³ Total Volume: VT = 1 + x³ Question 4 the range of this quadratic function y = -x² + 6x - 2 Answer Given function is: f(x) = -x² + 6x - 2 which is a quadratic function, whose graph is a parabola. To find the critical points of f(x). Let f'(x) = 0 -2x + 6 = 0 -2x = -6 => x = 3 is the only critical value. We can either use the knowledge of quadratics or take the second derivatives f''(x) = -2 => f''(3) = -2 < 0 So, at x=3, the function has maximum. And the maximum value is f(3) = -(3)² + 6(3) - 2 = -9 + 18 - 2 = 7 So, the range of the given quadratic function is [-∞, 7] Question 5 write the standard form of a quadratic equation with roots of -6 and 4 Answer (x - (-6)) * (x - 4) = 0 => (x + 6) * (x - 4) = 0 => x² - 4x + 6x - 24 = 0 => x² + 2x - 24 = 0