1.4 Quadratic Equations Solving a Quadratic Equation Completing the Square The Quadratic Formula 1.41.1 - 1 Quadratic Equation in One Variable An equation that can be written in the form ax2 + bx += c 0 where a, b, and c are real numbers with a ≠ 0, is a quadratic equation. The given form is called standard form. 1.4 - 2 Second-degree Equation A quadratic equation is a second-degree equation. This is an equation with a squared variable term and no terms of greater degree. x22=25, 4 xx + 4 −= 5 0, 3 xx2 = 4 − 8 1.4 - 3 Zero-Factor Property If a and b are complex numbers with ab = 0, then a = 0 or b = 0 or both. 1.4 - 4 Example 1 USING THE ZERO-FACTOR PROPERTY Solve 6xx2 += 73 Solution: 6xx2 += 73 6XX2 + 7 −= 30 Standard form ( 3xx− 1)( 2 += 3 ) 0 Factor. −= + = Zero-factor 3xx 1 0 or 2 3 0 property. 1.4 - 5 Example 1 USING THE ZERO-FACTOR PROPERTY Solve 6xx2 += 73 Solution: 3xx−= 1 0 or 2 + 3 = 0 Zero-factor property. 31x = or 23x = − Solve each equation. 1 3 x = or x = − 3 2 1.4 - 6 Square Root Property If x2 = k, then x= k or x= − k 1.4 - 7 Example 2 USING THE SQUARE ROOT PROPERTY Solve each quadratic equation. a. x 2 = 17 Solution: By the square root property, the solution set is {± 17} 1.4 - 8 Example 2 USING THE SQUARE ROOT PROPERTY Solve each quadratic equation. b. x 2 = −25 Solution: Since −=1,i the solution set of x2 = −25 is {±5.i} 1.4 - 9 Example 2 USING THE SQUARE ROOT PROPERTY Solve each quadratic equation. c. (x −= 4)2 12 Solution: Use a generalization of the square root property. 2 (x −= 4) 12 Generalized square x −=±4 12 root property. x =4 ± 12 Add 4. x =4 ± 23 12= 4 3 = 2 3 1.4 - 10 Solving A Quadratic Equation By Completing The Square To solve ax2 + bx + c = 0, by completing the square: Step 1 If a ≠ 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 Square half the coefficient of x, and add this square to both sides 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. 1.4 - 11 Example 3 USING THE METHOD OF COMPLETING THE SQUARE a = 1 Solve x2 – 4x –14 = 0 by completing the square. Solution Step 1 This step is not necessary since a = 1. Step 2 xx2 − 4 = 14 Add 14 to both sides. 2 1 2 Step 3 xx−=4++4 14 4 ()− 4 = 4; 2 add 4 to both sides. Step 4 (x −= 2)2 18 Factor; combine terms. 1.4 - 12 Example 3 USING THE METHOD OF COMPLETING THE SQUARE a = 1 Solve x2 – 4x –14 = 0 by completing the square. Solution Step 4 (x −= 2)2 18 Factor; combine terms. Step 5 x −=±2 18 Square root property. Take both Add 2. roots. x =2 ± 18 x =2 ± 32 Simplify the radical. The solution set is {2± 3 2.} 1.4 - 13 Example 4 USING THE METHOD OF COMPLETING THE SQUARE a ≠ 1 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution 9xx2 − 12 += 9 0 4 xx2 − +=10 Divide by 9. (Step 1) 3 4 xx2 −=−1 Add –1. (Step 2) 3 2 2 4 4414 44 xx−=+ −1+ − = ; add 3 9 9 23 99 1.4 - 14 Example 4 USING THE METHOD OF COMPLETING THE SQUARE a = 1 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution 2 2 4 4414 44 xx−=+ −1+ − = ; add 3 9 9 23 99 252 x −=− Factor, combine 39terms. (Step 4) 25 x − =±− Square root property 39 1.4 - 15 Example 4 USING THE METHOD OF COMPLETING THE SQUARE a = 1 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution 25 x − =±− Square root property 39 25 −=a ia xi−=± Quotient rule for 33 radicals 25 xi= ± Add ⅔. 33 1.4 - 16 Example 4 USING THE METHOD OF COMPLETING THE SQUARE a = 1 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution 25 xi= ± Add ⅔. 33 25 The solution set is ± i . 33 1.4 - 17 Quadratic Formula The solutions of the quadratic equation ax2 + bx + c = 0, where a ≠ 0, are −±b b2 −4 ac x = . 2a 1.4 - 18 Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Solve x2 – 4x = –2 Solution: xx2 −4 += 20 Write in standard form. Here a = 1, b = –4, c = 2 −±b b2 −4 ac x = Quadratic formula. 2a 1.4 - 19 Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Solve x2 – 4x = –2 Solution: −±b b2 −4 ac x = Quadratic formula. 2a −±()−−4 ()4()()4 2 −1 2 = 2(1) The fraction bar extends under –b. 1.4 - 20 Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Solve x2 – 4x = –2 Solution: −±()−−4 ()4()()4 2 −1 2 = 2(1) The fraction bar extends under –b. 4±− 16 8 = 2 4± 22 = 16−= 8 8 = 4 2 = 2 2 2 1.4 - 21 Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Solve x2 – 4x = –2 Solution: 4± 22 = 16−= 8 8 = 4 2 = 2 2 2 2(22± ) = Factor out 2 in the numerator. 2 Factor first, then divide. =22 ± Lowest terms. The solution set is {2± 2.} 1.4 - 22 Example 6 USING THE QUADRATIC FORMULA (NONREAL COMPLEX SOLUTIONS) Solve 2x2 = x – 4. Solution: 2 2xx−+= 40Write in standard form. −−( 1) ± ( − 1)2 − 4( 2)( 4 ) x = Quadratic formula; 2(2) a = 2, b = –1, c = 4 Use parentheses and 1±− 1 32 substitute carefully to = avoid errors. 4 1.4 - 23 Example 6 USING THE QUADRATIC FORMULA (NONREAL COMPLEX SOLUTIONS) Solve 2x2 = x – 4. Solution: 1±− 1 32 = 4 1± −31 x = −=1 i 4 1 31 The solution set is ± i . 44 1.4 - 24.