Solving Equations An equation in one variable is a statement in which two expressions, at least one of which containing the variable, are equal. The expressions are called the sides of the equation. Since an equation is a statement, it may be either true or false, depending on the value of the variable. The values of the variable that result in a true statement are called the solutions or roots of the equation. Examples: p 2 x2−4 2 x + 5 = 9 x + 5x = 2x − 2 x+1 = 0 x + 9 = 5 Sometimes equations can have more than one solution: x2−4 x+1 = 0 has x = −2 and x = 2 has solutions. One method for solving equations is to replace the original equation by a succession of equivalent equations. Equivalent Equations The following procedures result in equivalent equations. 1. Interchange the two sides of the equation: Replace 3 = x by x = 3 2. Simplify the sides of the equation by combing like terms, eliminating parentheses, and so on: Replace (x + 2) + 6 = 2x + (x + 1) by x + 8 = 3x + 1 3. Add or Subtract the same expression on both sides of the equation: Replace 3x − 5 = 4 by (3x − 5) + 5 = 4 + 5 4. Multiply or divide both sides of the equation by the same nonzero expression: 3x 6 Replace x−1 = x−1 , x 6= 1 3x 6 by x−1 ∗ (x − 1) = x−1 ∗ (x − 1) 5. If one side of the equation is 0 and the other side can be factored, then we may set each factor equal to 0: Replace x(x − 3) = 0 by x = 0 or x − 3 = 0 1 Example: 8x − (3x + 2) = 3x − 10 8x − 3x − 2 = 3x − 10 5x − 2 = 3x − 10 2x = −8 x = −4 Example: 2 1 1 3 p = 2 p + 3 4p = 3p + 2 p = 2 Example: (x + 2)(x − 3) = (x + 3)2 x2 − 3x + 2x − 6 = (x + 3)(x + 3) x2 − x − 6 = x2 + 6x + 9 x2 − x2 − x − 6x = 9 + 6 −7x = 15 −15 x = 7 Example: 5 4 9 y+2 − y−6 = y+4 5(y − 6)(y + 4) + 4(y + 2)(y + 4) = 9(y + 2)(y − 6) 5(y2 − 2y − 24) + 4(y2 + 6y + 8) = 9(y2 − 4y − 12) 5y2 − 10y − 120 + 4y2 + 24y + 32 = 9y2 − 36y − 108 14y − 88 = −36y − 108 50y = −20 2 y = 5 2 Solving Quadratic Equations Definition: A quadratic equation is an equation equivalent to one of the form: ax2 + bx + c = 0, where a; b; and c are real numbers and a 6= 0. A quadratic equation in this form is said to be in standard form. Once a quadratic equation is written in standard form, you can solve for x by: 1. factoring ax2 + bx + c into the product of two linear polynomials and then applying the zero product property, or 2. by using the square root method, or 3. by using the quadratic formula. Example 1: Solve the equation x2 − 4 = 0. () x2 = 4 p () x = ± 4 () x = 2 or x = −2 So the solution in set notation is x 2 {−2; 2g Example 2: Solve the equation x2 − 5x = 0. () x(x − 5) = 0 () x = 0 or x − 5 = 0 () x = 0 or x = 5. So the solution in set notation is x 2 f0; 5g Example 3: Solve the equation x2 + 5x = −6. () x2 + 5x + 6 = 0 () (x + 2)(x + 3) = 0 () x + 2 = 0 or x + 3 = 0 () x = −2 or x = −3. So the solution in set notation is x 2 {−2; −3g Example 4: Solve the equation x2 + 7x + 12 = 0. () (x + 3)(x + 4) = 0 () x + 3 = 0 or x + 4 = 0 () x = −3 or x = −4. So the solution in set notation is x 2 {−3; −4g 3 Example 5: Solve the equation 3x2 − 7x − 6 = 0. () (3x + 2)(x − 3) = 0 () 3x + 2 = 0 or x − 3 = 0 2 () x = − 3 or x = 3. 2 So the solution in set notation is x 2 {− 3 ; 3g Recall: For a quadratic function in standard form, the quadratic formula is p −b± b2−4ac x = 2a Example 6: Solve 3x2 − 5x + 1 = 0. () 3x2 + (−5)x + 1 = 0 ) a = 3, b = −5, c = 1 p p p 5± 52−4(3)(1) 5± 25−12 5± 13 So x = 2(3) = 6 = 6 . p p n 5− 13 5+ 13 o Thus x 2 6 ; 6 Note: If b2 − 4ac < 0, then the quadratic equation has no (real) solutions. 4.