Real Quadratic Formula If AWS
Real Quadratic Formula If 𝑓(𝑥 ) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 (𝑎, 𝑏, 𝑐 in ℝ and 𝑎 ≠ 0), then the roots of 𝑓(𝑥) are given by −𝑏 ± √𝑏2 − 4𝑎𝑐 𝑤= 2𝑎 Theorem (Roots of a real quadratic) Let 𝑓(𝑥 ) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 be a real quadratic (𝑎, 𝑏, 𝑐 in ℝ). Then 𝑓(𝑥) either has two real roots or a repeating real root or a pair of conjugate complex roots. Examples Find the roots of the following polynomials a.
𝑓 (𝑥 ) = 3𝑥 2 − 2𝑥 − 1 −(−2) ± √(−2)2 − 4(3)(−1) 𝑤= 2(3) 2 ± √4 + 12 = 6 2 ± √16 = 6 2±4 = 6 1 = 1 𝑜𝑟 − 3 two real roots
b.
𝑔(𝑥 ) = 𝑥 2 + 6𝑥 + 9
−6 ± √(6)2 − 4(1)(9) 𝑤= 2(1) −6 ± √0 = 2 = −3 a repeating real root *The root is “repeating” because if we factor this quadratic, we get 𝑔(𝑥 ) = (𝑥 + 3)2 , notice that the factor (𝑥 + 3) is raised to a power greater than 1, so the root −3 is a repeated root. We can say that it has multiplicity 2. c.
ℎ(𝑥 ) = 2𝑥 2 − 4𝑥 + 9 −(−4) ± √(−4)2 − 4(2)(9) 𝑤= 2(2) 4 ± √−56 𝑤= 4 4 ± √(−1)(4)(14) 𝑤= 4 4 ± √(−1)√4√14 𝑤= 4 4 ± 2𝑖 √14 𝑤= 4 𝑤 =1±𝑖
√14 2
conjugate complex roots
Theorem (Complex roots) For any complex number 𝑤, there exists a real quadratic 𝑓(𝑥) with roots 𝑤 and 𝑤 ̅. (i.e. given any pair of conjugate numbers, there is a quadratic that “fits” these roots)
Theorem Suppose 𝑓(𝑥 ) = 𝑥 2 + 𝑞𝑥 + 𝑟, where 𝑞 and 𝑟 are complex numbers. Let 𝑢 and 𝑤 be roots of this quadratic, then 𝑞 = −(𝑢 + 𝑤) and 𝑟 = 𝑢𝑤 Examples 1.
Find a quadratic that has 𝑤 = 2 + 3𝑖 as a root. Given the conjugate pair 𝑤 = 2 + 3𝑖 and 𝑤 ̅ = 2 − 3𝑖, there exists a real quadratic that has these as roots. 0 = (𝑥 − (2 + 3𝑖 ))(𝑥 − (2 − 3𝑖 )) 0 = 𝑥 2 − 𝑥 (2 − 3𝑖 ) − (2 + 3𝑖 )𝑥 + (2 + 3𝑖 )(2 − 3𝑖 ) 0 = 𝑥 2 − 2𝑥 + 3𝑖𝑥 − 2𝑥 − 3𝑖𝑥 + 4 − 6𝑖 + 6𝑖 + 9 0 = 𝑥 2 − 4𝑥 + 13 Therefore, the real quadratic is 𝑓 (𝑥 ) = 𝑥 2 − 4𝑥 + 13.
2.
Given that 2 + 𝑖 is a root of 𝑓 (𝑥 ) = 𝑥 2 − 2𝑥 + (1 − 2𝑖 ), find the other root of 𝑓. According to the theorem, the quadratic is 𝑓 (𝑥 ) = 𝑥 2 + 𝑞𝑥 + 𝑟 where 𝑞 = −(𝑢 + 𝑤) and 𝑟 = 𝑢𝑤 with roots 𝑢 and 𝑤. If we let 𝑢 = 2 + 𝑖 and 𝑤 = 𝑎 + 𝑏𝑖, then −2 = −(2 + 𝑖 + 𝑎 + 𝑏𝑖 ) 2 = 2 + 𝑎 + 𝑖 (1 + 𝑏) Equating the real and imaginary parts, 𝑎 = 0 and 𝑏 = −1 We check if these values work for the 𝑟 equation: (1 − 2𝑖 ) = (2 + 𝑖 )(0 − 1𝑖 ) (1 − 2𝑖 ) = (−2𝑖 + 1) Indeed, 𝑎 = 0, 𝑏 = −1 the other root is 𝑤 = −𝑖
Complex Quadratic Formula It turns out that we can use the same quadratic formula for real and complex polynomials. i.e. If 𝑓 (𝑥 ) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 (𝑎, 𝑏, 𝑐 in ℂ and 𝑎 ≠ 0), then the roots of 𝑓(𝑥) are given by −𝑏 ± √𝑏2 − 4𝑎𝑐 𝑢= 2𝑎 How to find square roots of complex numbers? Use DeMoivre’s Law (see section 8)
Practice Problems 1. Find the roots of the quadratic equation 0 = 𝑧 2 − 2𝑧 + 5.
𝑓 (𝑥 ) = 3𝑥 2 − 2𝑥 − 1 −(−2) ± √(−2)2 − 4(3)(−1) 𝑤= 2(3) 2 ± √4 + 12 = 6 2 ± √16 = 6 2±4 = 6 1 = 1 𝑜𝑟 − 3 two real roots
b.
𝑔(𝑥 ) = 𝑥 2 + 6𝑥 + 9
−6 ± √(6)2 − 4(1)(9) 𝑤= 2(1) −6 ± √0 = 2 = −3 a repeating real root *The root is “repeating” because if we factor this quadratic, we get 𝑔(𝑥 ) = (𝑥 + 3)2 , notice that the factor (𝑥 + 3) is raised to a power greater than 1, so the root −3 is a repeated root. We can say that it has multiplicity 2. c.
ℎ(𝑥 ) = 2𝑥 2 − 4𝑥 + 9 −(−4) ± √(−4)2 − 4(2)(9) 𝑤= 2(2) 4 ± √−56 𝑤= 4 4 ± √(−1)(4)(14) 𝑤= 4 4 ± √(−1)√4√14 𝑤= 4 4 ± 2𝑖 √14 𝑤= 4 𝑤 =1±𝑖
√14 2
conjugate complex roots
Theorem (Complex roots) For any complex number 𝑤, there exists a real quadratic 𝑓(𝑥) with roots 𝑤 and 𝑤 ̅. (i.e. given any pair of conjugate numbers, there is a quadratic that “fits” these roots)
Theorem Suppose 𝑓(𝑥 ) = 𝑥 2 + 𝑞𝑥 + 𝑟, where 𝑞 and 𝑟 are complex numbers. Let 𝑢 and 𝑤 be roots of this quadratic, then 𝑞 = −(𝑢 + 𝑤) and 𝑟 = 𝑢𝑤 Examples 1.
Find a quadratic that has 𝑤 = 2 + 3𝑖 as a root. Given the conjugate pair 𝑤 = 2 + 3𝑖 and 𝑤 ̅ = 2 − 3𝑖, there exists a real quadratic that has these as roots. 0 = (𝑥 − (2 + 3𝑖 ))(𝑥 − (2 − 3𝑖 )) 0 = 𝑥 2 − 𝑥 (2 − 3𝑖 ) − (2 + 3𝑖 )𝑥 + (2 + 3𝑖 )(2 − 3𝑖 ) 0 = 𝑥 2 − 2𝑥 + 3𝑖𝑥 − 2𝑥 − 3𝑖𝑥 + 4 − 6𝑖 + 6𝑖 + 9 0 = 𝑥 2 − 4𝑥 + 13 Therefore, the real quadratic is 𝑓 (𝑥 ) = 𝑥 2 − 4𝑥 + 13.
2.
Given that 2 + 𝑖 is a root of 𝑓 (𝑥 ) = 𝑥 2 − 2𝑥 + (1 − 2𝑖 ), find the other root of 𝑓. According to the theorem, the quadratic is 𝑓 (𝑥 ) = 𝑥 2 + 𝑞𝑥 + 𝑟 where 𝑞 = −(𝑢 + 𝑤) and 𝑟 = 𝑢𝑤 with roots 𝑢 and 𝑤. If we let 𝑢 = 2 + 𝑖 and 𝑤 = 𝑎 + 𝑏𝑖, then −2 = −(2 + 𝑖 + 𝑎 + 𝑏𝑖 ) 2 = 2 + 𝑎 + 𝑖 (1 + 𝑏) Equating the real and imaginary parts, 𝑎 = 0 and 𝑏 = −1 We check if these values work for the 𝑟 equation: (1 − 2𝑖 ) = (2 + 𝑖 )(0 − 1𝑖 ) (1 − 2𝑖 ) = (−2𝑖 + 1) Indeed, 𝑎 = 0, 𝑏 = −1 the other root is 𝑤 = −𝑖
Complex Quadratic Formula It turns out that we can use the same quadratic formula for real and complex polynomials. i.e. If 𝑓 (𝑥 ) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 (𝑎, 𝑏, 𝑐 in ℂ and 𝑎 ≠ 0), then the roots of 𝑓(𝑥) are given by −𝑏 ± √𝑏2 − 4𝑎𝑐 𝑢= 2𝑎 How to find square roots of complex numbers? Use DeMoivre’s Law (see section 8)
Practice Problems 1. Find the roots of the quadratic equation 0 = 𝑧 2 − 2𝑧 + 5.
