Questions and Answers #12 Quadratics

Questions and Answers Sheet 12 Quadratics Question #1 Solve 𝑥 4 + 16 = 0. Express the roots in rectangular form . Answer: The given expression is 𝑥 4 + 16 = 0 Adding and subtracting 8𝑥 2 , we get 𝑥 4 − 8𝑥 2 + 8𝑥 2 + 16 = 0 (𝑥 2 + 4)2 − 8𝑥 2 = 0 2 (𝑥 2 + 4)2 − (2√2𝑥) = 0 (𝑥 2 + 4 + 2√2𝑥)(𝑥 2 + 4 − 2√2𝑥) = 0 (∵ 𝑎2 − 𝑏 2 = (𝑎 + 𝑏)(𝑎 − 𝑏)) Solving the two quadratics separately. (𝑥 2 + 4 + 2√2𝑥) = 0 By quadratic formula 𝑥= = = −2√2±√8−4×4 2 −2√2±√−8 2 −2√2±2√2𝑖 2 = −√2 ± √2𝑖 Now , solving another quadratic equation, we get (𝑥 2 + 4 − 2√2𝑥) = 0 By quadratic formula 𝑥= = = 2√2±√8−4×4 2 2√2±√−8 2 2√2±2√2𝑖 2 = √2 ± √2𝑖 Hence the roots of the given expression in rectangular form are −√2 + √2𝑖 −√2 − √2𝑖 √2 + √2𝑖 √2 − √2𝑖 Question #2 What would be the best method to solve −3𝑥 2 + 𝑥 + 1 = 0? Answer: Given: −3𝑥 2 + 12𝑥 + 1 = 0 This is in the form: 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 with 𝑎 = −3, 𝑏 = 12 and 𝑐 = 1 In order to decide which method to use, we can first examine the discriminant: Δ = 𝑏 2 − 4𝑎𝑐 = 122 − 4(−3)(1) = 144 + 12 = 156 Since Δ > 0 we can tell that the quadratic has 2 distinct real solutions, but because it is not a perfect square we can also tell that those solutions are irrational. So there is no rational factorisation and we can discount the use of an AC method or similar. We can use the quadratic formula or we can complete the square. If we choose to use the quadratic formula, then we find: 𝑥= = = = −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 −𝑏±√Δ 2𝑎 −(12)±√156 2(−3) −12±2√39 −6 √39 =2± 3 When completing the square, note that b=12 is already divisible by a=-3 and by 2. So to do most our arithmetic using integers, we can premultiply by -3 to make the leading term square: 0 = −3(−3𝑥 2 + 12𝑥 + 1) = 9𝑥 2 − 36𝑥 − 3 = (3𝑥)2 − 2(3𝑥)(6) + (6)2 − 39 2 = (3𝑥 − 6)2 − (√39) = ((3𝑥 − 6) − √39) ((3𝑥 − 6) + √39) = (3𝑥 − 6 − √39)(3𝑥 − 6 + √39) So: 3𝑥 = 6 ± √39 𝑥 =2± √39 3 Question #3 How do you solve −4𝑥 2 + 4𝑥 = 9 ? Answer: For a general quadratic 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, you know that the solutions are given by 𝑥1,2 = −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 First of all, we rewrite the equation as −4𝑥 2 + 4𝑥 − 9 = 0, which (if you prefer) can be also written as 4𝑥 2 − 4𝑥 + 9 = 0 In this final case, 𝑎 = 4, 𝑏 = −4, 𝑐 = 9 . Substituing this values into the solving formula, we get 𝑥1,2 = 4±√16−4⋅4⋅9 2⋅4 The discriminant 𝑏 2 − 4𝑎𝑐 is negative, and thus there are no real solutions. If you're interested in complex solutions, you get 𝑥1,2 = 4±𝑖√128 8 = 4±8𝑖√2 8 which yelds 1 𝑥1 = + 𝑖√2. 2 1 𝑥2 = − 𝑖√2. 2 Question #4 Which method do you use to solve the quadratic equation 81𝑥 2 + 1 = 0 ? Answer: I would try to isolate x. 81𝑥 2 + 1 = 0 by subtracting 1, ⇒ 81𝑥 2 = −1 by dividing by 81, 1 ⇒ 𝑥2 = − 81 Since 𝑥 2 can never be negative for a real number x, the equation has no real solutions; however, if you are looking for its complex solutions, then by taking the square-root, ⇒ 𝑥 = ±√− 1 81 =± 𝑖 9 Question #5 How do you solve for x in 𝑥 2 − 6𝑥 + 4 = 0? Answer: I would find the roots for this equation using the quadratic formula 𝑥 2 − 6𝑥 + 4 = 0 Use quadratic formula with a=1, b=-6, c=4 . −𝑏 ± √𝑏2−4𝑎𝑐 2 𝑎 𝑥 = −(−6) ± 𝑥 =6± √(−6)2−4(1)(4) 2 (1) √20 2 Answer: 𝑥 = 5.23606797749979, 0.7639320225002102 Question #6 How do you solve 𝑥 2 − 6𝑥 − 16 = 0 by factoring? Answer: Factoring a polynomial p(x) means to find its roots 𝑥𝑖 , and divide the polynomial by the term (𝑥 − 𝑥𝑖 ) , writing p(x) as (𝑥 − 𝑥𝑖 )𝑞(𝑥) , and then iterating the process searching the roots of q(x) (which of course are roots of p(x), too. The fundamental theorem of algebra states that you can always write a polynomial of degree k as the product of k linear factors (𝑥 − 𝑥𝑖 ), i.e., every polynomial of degree k has k complex roots. These roots can be complex but not real, and in that case, you cannot simplify the polynomial in R. In your case, the solutions are easy to find, without doing any calculation: you can remember the rule which states that, if you have a quadratic of the form 𝑥 2 − 𝑠𝑥 + 𝑝, the product of the solutions equals p and the sum of the solutions equals s. So, we have to find two numbers which sum up to -6, and that multiplied one for the other give -16. You can easily convince yourself that these two numbers are -2 and 8. If -2 and 8 are roots of our polynomial, for what we said above the following must hold: 𝑥 2 − 6𝑥 − 16 = (𝑥 + 2)(𝑥 − 8) Question #7 How do you solve by factoring and using the principle of zero products 𝑥 2 + 7𝑥 + 6 = 0 ? Answer: = −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 #8 How do you solve 𝑥 2 − 9𝑥 + 14 = 0? Answer: Solve 𝑥 2 − 9𝑥 + 14 = 0. The answer is 𝑥 = 7 and 𝑥 = 2. Factor the left side of the equation. (𝑥 − 7)(𝑥 − 2) = 0 Solve each term in the product separately. x-7=0 Add 7 to both sides. x=7 x-2=0 Add 2 to both sides. x=2 x=7 72 − 9(7) + 14 = 0 49-63+14=0 x=2 22 − 9(2) + 14 = 0 4-18+14=0 Question #9 How do you use algebra to solve (x-1)(x+2)=18 ? Answer: We know that a if a product is 0, then at least one of the factors must be 0. But, that's not what we have here. Let's try to make it what we have: (x-1)(x+2)=18 (x-1)(x+2)-18=0 [𝑥 2 + 𝑥 − 2] − 18 = 0 (I multiplied (x-1)(x+2)) 𝑥 2 + 𝑥 − 20 = 0 (Simplify) (x+5)(x-4)=0 (Factor the NEW expression) 𝑥 + 5 = 0 or 𝑥 − 4 = 0 It is clear that there are 2 solutions: -5 is a solution and 4 is a solution. Question #10 How do solve 𝑛2 − 10𝑛 + 22 = −2 ? Answer: 𝑛2 − 10𝑛 + 22 = −2 is equivalent to 𝑛2 − 10𝑛 + 24 = 0 We would like to factor the left hand size to get something of the form (𝑛 + 𝑎)(𝑛 + 𝑏) = 𝑛2 − 10𝑛 + 24 Since the final term is positive, a and b must have the same sign and since the coefficient of x is negative they must both be negative. So we are looking for negative values a and b such that a+b=-10 and ab=+24 Assuming integer solutions there are very few possibilities. By trial and error we can derive (𝑛 − 6)(𝑛 − 4) = 𝑛2 − 10𝑛 + 24 = 0 which implies that 𝑛 = 6 or 𝑛 = 4