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Section A.5 Solving Equations Solving Linear Equations Solving Rational

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Section A.5 Solving Equations Solving Linear Equations Solving Rational Lecture 5: Section A.5 Solving Equations Solving linear equations Solving rational equations Extraneous solutions Solving quadratic equations Completing the square Quadratic formula Discriminant Solving equations with radicals or rational exponents Solving absolute value equations L5 - 1 Def. An equation in x is a statement that two algebraic expressions are equal. To solve an equation is to find all values of x for which the equation is true. Such values of x are solutions (or roots, zeros) of the equation. NOTE: If an equation has no solution, the solution set is empty, written as ;. ex. Solve the equation jx − 1j = −2 Linear Equations Def. A linear equation in one variable x is an equation of the form ax + b = 0 where a and b are real numbers with a =6 0. L5 - 2 To solve a linear equation: 1. Remove all parentheses and simplify each side of the equation as much as possible. 2. Rewrite the equation by isolating the variable: variable terms on one side, numbers on the other. 3. Solve for the variable and check your solution. ex. Solve 6(x − 1) + 4 = 3(7x + 1) NOTE: A linear equation has exactly one solution. Solve a linear equation with fractions, multiply both sides by LCD to clear the fraction. x 1 ex. + 1 = (x − 6) 2 4 L5 - 3 To solve rational equations that lead to linear equations 1. Find the domain of the variable. 2. Clear the equation of fractions by multiplying both sides by LCM of the denominator. 3. Solve for the variable. 4. Choose the solutions that are in the domain. 3 1 7 ex. Solve: = + x − 2 x − 1 x2 − 3x + 2 L5 - 4 2x 4 3 ex. Solve: = − x2 − 4 x2 − 4 x + 2 NOTE: We call x = 2 an extraneous solution. NOTE: An equation with a single fraction on each side can be cleared of denominators by cross multiplying. 3 1 ex. = x − 2 x − 1 L5 - 5 Quadratic Equations Def. A quadratic or second-degree equation is an equation that can be written in the general form ax2 + bx + c = 0 where a, b and c are real numbers and a =6 0. To solve a quadratic equation: 1. By Factoring: Use the Zero-Factor property: If ab = 0, then ex. Solve: 3x2 + 18x = 21 NOTE: The right side of the equation has to be 0 before factoring. L5 - 6 ex. Solve: (x + 3)(x − 4) = 8 2. By Square Root Principle: p If u2 = c, where c ≥ 0, then u = ± c. ex. Solve: (x + 2)2 = 5 L5 - 7 3. By Completing the Square of ax2 + bx + c = 0; a =6 0 1) Make sure a = 1. If not, divide each term by a. 2) Move the constant to the right side of the equation. 3) Complete the square by adding the square of one- half of the coefficient of x to each side of the equation. 4) Factor the left side as a perfect square. 5) Solve for x using the Square Root Principle. ex. Solve by completing the square: 3x2−4x−2 = 0 L5 - 8 4. By Quadratic Formula: If ax2 + bx + c = 0, a =6 0, then p −b ± b2 − 4ac x = 2a Proof. Complete the square. ex. Solve using the quadratic formula: 1) 2x2 + 8x + 8 = 0 2) 2x(3 − x) = 3 3) x2 + 2x + 2 = 0 L5 - 9 Def. The quantity b2 − 4ac is the discriminant of the quadratic equation. NOTE: The equation ax2 + bx + c = 0, 1) If b2−4ac > 0, then the equation has two distinct real number solutions. 2) If b2−4ac = 0, then the equation has one repeated real number solution. 3) If b2 − 4ac < 0, then the equation has no real number solutions. Practice. Use the discriminant to determine the number of real solutions of the equations 1) 2x2 + 9x + 1 = 0 2) 2x2 + 8x + 8 = 0 )todsic elsltos2 n eetdra solution real repeated one 2) solutions real distinct two 1) Answer. L5 - 10 Equations with Radicals or Rational Exponents To find the real solution of an equation with radicals: 1. Simplify the equation if possible. 2. Isolate the most complicated radical on one side of the equation. 3. Raise both sides of the equation to the index of the radical to eliminate the radical. 4. Check for extraneous solution, since raise both sides to an even power may add extraneous solutions. p ex. Solve: 3 + 3x + 1 = x L5 - 11 ex. Solve: (x + 3)2=3 = 4 p p ex. Solve: 3x + 1 − x − 1 = 2 L5 - 12 Absolute Value Equations Recall the definition: jaj = ex. Solve each equation: 1) 4 + j3 − xj = 2 2) jx2 − 6j = x NOTE: If the equation involves rational expres- sions, radicals, rational exponents or absolute values, you must check your solution(s) in the original equa- tion. L5 - 13 Practice. 1) Solve 16(x + 1)2 + 8(x + 1) + 1 = 0 (Hint: Use substitution u = x + 1) 2) Solve: x3 = 6x2 − 9x 3) Solve: x6 − 6x3 + 9 = 0 4) Solve: x3 − 3x2 − 3x + 9 = 0 5) Solve: (3x + 1)−1=2 + 2(3x + 1)1=2 = 0 6) Solve: (x − 1)2=3 + (x − 1)1=3 − 12 = 0 3 28 63, − 6) solution no 5) 4 3 ; 3 ; 3 − 3 4) 3 0, 3) 2) − 1) Answer. 3 5 p p p L5 - 14.
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