Section 1.4 Quadratic Equations Definition: An equation that can be written in the form ππ₯ 2 + ππ₯ + π = 0 where a, b, and c are real numbers with a β 0, is quadratic equation. The given form is called standard form. 1. Solving by Factoring: To solve a quadratic equation by factoring, rewrite the equation, if necessary, so that one side is equal to 0 and use the Zero-Product Property. Zero-Product Property: ππ = 0 if and only if π = 0 or π = 0. Example 1: Solve the following equations by factoring. a) π₯ 2 β 3π₯ = β2 b) π₯ 2 + 3π₯ = 18 c) 4π₯ 2 β 4π₯ = β1 Section 1.4 Quadratic Equations d) 6π₯ 2 + 7π₯ = 3 1|Page Square root Property: If π₯ 2 = π, then π₯ = βπ or π₯ = ββπ Example 2: Solve each quadratic equation. a) π₯ 2 = 17 c) (π₯ β 4)2 = 1 b) 49π₯ 2 + 36 = 0 d) (2π₯ β 1)2 = 12 2. Solving by Completing the Square: To solve ππ₯ 2 + ππ₯ + π = 0 with a β 0, by completing the square, use these steps. Step 1 If a β 1, divide both sides of the equation by a. Step 2 Rewrite the equation so that the constant term is alone on one side of the equality symbol. Step 3 Take a half of the coefficient of π₯, and then square the result. Now add this square to each side of the equation. Step 4 Factor the resulting trinomial as a perfect square and combine like terms on the other side. Step 5 Use the square root property to complete the solution. Example 3: Solve π₯ 2 β 4π₯ β 14 = 0 by completing the square. Step 1 Step 2 Step 3 Step 4 Step 5 Section 1.4 Quadratic Equations 2|Page Example 4: Solve 9π₯ 2 β 12π₯ + 9 = 0 by completing the square. 3. Solving by the Quadratic Formula: The solution of the quadratic equation ππ₯ 2 + ππ₯ + π = 0 with a β 0 are given by the quadratic formula. βπ ± βπ 2 β 4ππ π₯= 2π Example 5: Find all solutions of the following each quadratic equation: a) π₯ 2 β 4π₯ = β2 b) 2π₯ 2 = π₯ β 4 Example 6: Solve for the specified variable. Use ± when taking square roots. Section 1.4 Quadratic Equations 3|Page a) A = ππ2 4 , for π b) ππ‘ 2 β π π‘ = π (π β 0), for π‘ The quantity under the radical in the quadratic formula, π 2 β 4ππ, is called the discriminant. βπ ± βπ 2 β 4ππ π₯= 2π Example 7: Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. a) 5π₯ 2 + 2π₯ β 4 = 0 b) π₯ 2 β 10π₯ = β25 c) 2π₯ 2 β π₯ + 1 = 0 Section 1.4 Quadratic Equations 4|Page
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