9-9 The Qudrtic Formul nd the Discriminnt Objectives Solve qudrtic equtions by using the Qudrtic Formul. Determine the number of solutions of qudrtic eqution by using the discriminnt. Vocbulry discriminnt Why lern this? You cn use the discriminnt to determine whether the weight in crnivl strength test will rech certin height. (See Exercise.) In the previous lesson, you completed the squre to solve qudrtic equtions. If you complete the squre of x + bx + c = 0, you cn derive the Qudrtic Formul. The Qudrtic Formul cn be used to solve ny qudrtic eqution. Numbers Algebr x + 6x + 1 = 0 _ x + 6_ x + 1_ = 0_ Divide both sides by. x + bx + c = 0, 0 _ x + b_ x + c_ = 0_ x + 3x + 1 _ = 0 x + b _ x + c _ = 0 x + 3-1_ Subtrct c from both sides. x + b _ - c_ x + 3x + ( 3_ ) =- 1 _ + ( 3_ ) ( x + 3_ ) = 9_ - 1 _ Complete the squre. Fctor nd simplify. x + b _ x + ( b_ ) =- c _ + ( b_ ) ( x + b b ) = - _ c To dd frctions, you need common denomintor. b - c = b - c = b ( ) - c = b - c _ ( x + _ 3 ) = _ 9 - _ ( x + _ 3 ) 7_ = x + _ 3 = ± _7-3_ ± _7 Use common denomintors. Tke squre roots. Subtrct b from both sides. ( x + _ ) b = b _ - _ c _ ( x + _ ) b = b - c x + _ b = ± b - c b_ - ± b - c -3 _7 ± -b ± b - c The Qudrtic Formul The solutions of x + bx + c = 0, where 0, re -b ± b - c. 670 Chpter 9 Qudrtic Functions nd Equtions
EXAMPLE 1 Using the Qudrtic Formul You cn grph the relted qudrtic function to see if your solutions re resonble. Solve using the Qudrtic Formul. A x + 3x - 5 = 0 x + 3x + (-5) = 0 Identify, b, nd c. -b ± b - c Use the Qudrtic Formul. -3 ± 3 - () (-5) () -3 ± 9 - (-0) -3 ± _ 9 = _ -3 ± 7 _ -3 + 7 or _ -3-7 1 or -_ 5 B x - 3 1 x + (-) x + (-3) = 0 -(-) ± (-) - (1)(-3) (1) ± - (-1) _ ± 16 = _ ± _ + or _ - 3 or -1 Substitute for, 3 for b, nd -5 for c. Write s two equtions. Write in stndrd form. Identify, b, nd c. Substitute 1 for, - for b, nd -3 for c. Write s two equtions. Solve using the Qudrtic Formul. 1. -3 x + 5x + = 0 1b. - 5 x = -9x Mny qudrtic equtions cn be solved by grphing, fctoring, tking the squre root, or completing the squre. Some cnnot be esily solved by ny of these methods, but you cn use the Qudrtic Formul to solve ny qudrtic eqution. EXAMPLE Using the Qudrtic Formul to Estimte Solutions Solve x - x - = 0 using the Qudrtic Formul. -(-) ± (-) - (1)(-) (1) ± - (-16) = _ ± 0 _ + 0 or _ - 0 Use clcultor: x 3. or x -1.. Check resonbleness. Solve x - 8x + 1 = 0 using the Qudrtic Formul. 9-9 The Qudrtic Formul nd the Discriminnt 671
If the qudrtic eqution is in stndrd form, the discriminnt of qudrtic eqution is b - c, the prt of the eqution under the rdicl sign. Recll tht qudrtic equtions cn hve two, one, or no rel solutions. You cn determine the number of solutions of qudrtic eqution by evluting its discriminnt. Eqution x - x + 3 = 0 x + x + 1 = 0 x - x + = 0 Discriminnt = 1, b = -, c = 3 Grph of Relted Function b - c (-) - (1) (3) 16-1 The discriminnt is positive. Notice tht the relted function hs two x-intercepts. = 1, b =, c = 1 b - c - (1) (1) - 0 The discriminnt is zero. Notice tht the relted function hs one x-intercept. = 1, b = -, c = b - c (-) - (1) () - 8 - The discriminnt is negtive. Notice tht the relted function hs no x-intercepts. Number of Solutions two rel solutions one rel solution no rel solutions The Discriminnt of Qudrtic Eqution x + bx + c = 0 If b - c > 0, the eqution hs two rel solutions. If b - c = 0, the eqution hs one rel solution. If b - c < 0, the eqution hs no rel solutions. EXAMPLE 3 Using the Discriminnt Find the number of rel solutions of ech eqution using the discriminnt. A 3 x + 10x + = 0 = 3, b = 10, c = b - c 10 - (3) () 100-76 b - c is positive. There re two rel solutions. B 9 x - 6x + 1 = 0 = 9, b = -6, c = 1 b - c (-6) - (9) (1) 36-36 0 b - c is zero. There is one rel solution. C x + x + 1 = 0 = 1, b = 1, c = 1 b - c 1 - (1) (1) 1 - -3 b - c is negtive. There re no rel solutions. Find the number of rel solutions of ech eqution using the discriminnt. 3. x - x + 3 = 0 3b. x + x + = 0 3c. x - 9x + = 0 67 Chpter 9 Qudrtic Functions nd Equtions
The height h in feet of n object shot stright up with initil velocity v in feet per second is given by h = -16 t + vt + c, where c is the beginning height of the object bove the ground. 0 ft 15 ft EXAMPLE Physics Appliction If the object is shot stright up from the ground, the initil height of the object bove the ground equls 0. A weight 1 foot bove the ground on crnivl strength test is shot stright up with n initil 5 ft velocity of 35 feet per second. Will it ring the bell t the top of the pole? Use the discriminnt to explin your nswer. h = -16 t + vt + c 0 = -16t + 35t + 1 Substitute 0 for h, 35 for v, nd 1 for c. 0 = -16t + 35t + (-19) Subtrct 0 from both sides. b - c Evlute the discriminnt. 35 - (-16)(-19) = 9 Substitute -16 for, 35 for b, nd -19 for c. The discriminnt is positive, so the eqution hs two solutions. The weight will rech height of 0 feet so it will ring the bell. 10 ft. Wht if? Suppose the weight is shot stright up with n initil velocity of 0 feet per second. Will it ring the bell? Use the discriminnt to explin your nswer. There is no one correct wy to solve qudrtic eqution. Mny qudrtic equtions cn be solved using severl different methods. EXAMPLE 5 Solving Using Different Methods Solve x + 7x + 6 = 0. Method 1 Solve by grphing. y = x + 7x + 6 Write the relted qudrtic function nd grph it. The solutions re the x-intercepts, -6 nd -1. Method Solve by fctoring. x + 7x + 6 = 0 (x + 6) (x + 1) = 0 x + 6 = 0 or x - 1 = 0-6 or -1 Method 3 Solve by completing the squre. x + 7x + 6 = 0 x + _ 7-6 x + 7x + 9 _ 9 = -6 + ( x + _ 7 ) = _ 5 x + _ 7 = ± _ 5 x + _ 7 = _ 5 or x + _ 7 = - _ 5-1 or -6 Fctor. Use the Zero Product Property. Solve ech eqution. Add ( b ) to both sides. Fctor nd simplify. Tke the squre root of both sides. Solve ech eqution. 9-9 The Qudrtic Formul nd the Discriminnt 673
Method Solve using the Qudrtic Formul. 1 x + 7x + 6 = 0 Identify, b, nd c. -7 ± 7 - (1) (6) (1) -7 ± 9 - -7 = _ ± 5 _ -7 + 5 or _ -7-5 -1 or -6 Substitute 1 for, 7 for b, nd 6 for c. = _ -7 ± 5 Write s two equtions. Solve ech eqution. Solve. 5. x + 7x + 10 = 0 5b. -1 + x = 5x 5c. x + x - 1 = 0 Notice tht ll of the methods in Exmple 5 produce the sme solutions, -1 nd -6. The only method you cnnot use to solve x + 7x + 6 = 0 is using squre roots. Sometimes one method is better for solving certin types of equtions. The tble below gives some dvntges nd disdvntges of the different methods. Methods of Solving Qudrtic Equtions METHOD ADVANTAGES DISADVANTAGES Grphing Alwys works to give pproximte solutions Cn quickly see the number of solutions Cnnot lwys get n exct solution Fctoring Using squre roots Completing the squre Using the Qudrtic Formul Good method to try first Strightforwrd if the eqution is fctorble Quick when the eqution hs no x-term Alwys works Alwys works Cn lwys find exct solutions Complicted if the eqution is not esily fctorble Not ll qudrtic equtions re fctorble. Cnnot esily use when there is n x-term Sometimes involves difficult clcultions Other methods my be esier or less time consuming. Solving Qudrtic Equtions No mtter wht method I use, I like to check my nswers for resonbleness by grphing. Binh Phm Johnson High School I used the Qudrtic Formul to solve x - 7x - 10 = 0. I found tht x -1.09 nd x.59. Then I grphed y = x - 7x - 10. The x-intercepts ppered to be close to -1 nd.5, so I knew my solutions were resonble. 67 Chpter 9 Qudrtic Functions nd Equtions