Tiêu đề 6 Solving Equations.pdf ✅

x − 1 = ±√ 5 3 = ± √15 3 x = 1 ± √15 3 x = 3 ± √15 3 1.1.4.Quadratic Formula Derivation of the formula Quadratic Equation ax 2 + bx + c = 0 Divide both sides by a x 2 + b a x + c a = 0 Transpose the constant to the other side x 2 + b a x = − c a Complete the square by adding (− b 2a ) 2 = b 2 4a2 on both sides x 2 + b a x + b 2 4a 2 = b 2 − 4ac 4a 2 Factor the left-hand side (x + b 2a ) 2 = b 2 − 4ac 4a 2 Extract the square root x + b 2a = ± √b 2 − 4ac 2a Solve for x x = −b ± √b 2 − 4ac 2a - Example: 3x 2 − 5x − 1 = 0 x = −(−5) ± √(−5) 2 − 4(3)(−1) 2(3) x = 5 ± √37 6

2. Higher Degree Polynomial Equations 2.1. By Polynomial Division powered by the Factor Theorem and Rational Root Theorem • Rational Root Theorem is a theorem used to determine the first few guesses for a solution in a polynomial equation. • Factor Theorem is a theorem used to test whether a guess in a polynomial equation is a root. - Example: Solve for all values of x in the equation x 3 − 2x 2 − 5x + 6 = 0 [SOLUTION] From the rational root theorem, the possible roots are x = ±1, ±2, ±3, ±6. Try x = 1 by using division (the suggested method is the synthetic division) 1 1 −2 −5 6 1 −1 −6 1 −1 −6 0 Since the remainder is 0, then by Factor Theorem, x − 1 is a factor of the left-hand side of the equation. (x − 1)(x 2 − x − 6) = 0 Now, fully factor the quadratic factor by elementary methods (x − 1)(x − 3)(x + 2) = 0 { x − 1 = 0 x − 3 = 0 x + 2 = 0 → x = 1, 3, −2 • This method aims to factor the polynomial until all remaining factors are quadratic. 2.2. By the use of the Calculator (Equation mode) The process is similar to that of solving quadratic equations using a calculator. • In this discussion, the calculator used is limited to quartic equations. • One advantage of this method is the possibility of giving all roots regardless of nature (real or imaginary). • This method is limited depending on the degree of the equation and the capacity of the calculator.