9.4 Quadratics - Quadratic Formula

1 - Quadratic FormulaObjective: Solve Quadratic equations by using the Quadratic general from of a Quadratic isax2+bx+c= 0. We will now solve this for-mula forxby completing the squareExample +bc+c= 0 Separate constant from variables c cSubtractcfrom both sidesax2+bx= cDivide each term byaaaax2+bax= caFind the number that completes the square(12 ba)2=(b2a)2=b24a2 Add to both sides,b24a2 ca(4a4a)=b24a2 4ac4a2=b2 4ac4a2 Get common denominator on rightx2+bax+b24a2=b24a2 4ac4a2=b2 4ac4a2 Factor(x+b2a)2=b2 4ac4a2 Solve using the even root property(x+b2a)2 = b2 4ac4a2 Simplify rootsx+b2a= b2 4ac 2aSubtractb2afrom both sidesx= b b2 4ac 2aOur SolutionThis solution is a very important one to us. As we solved a general equation bycompleting the square, we can use this Formula to solve any Quadratic we identify whata, b,andcare in the Quadratic , we can substitute those1values intox= b b2 4ac 2aand we will get our two solutions. This Formula isknown as the Quadratic fromulaQuadratic Formula :ifax2+b x+c= 0thenx= b b2 4ac 2aWorld View Note:Indian mathematician Brahmagupta gave the first explicitformula for solving Quadratics in 628.

2 However, at that timemathematics was notdone with variables and symbols, so the Formula he gave was, To the absolutenumber multiplied by four times the square, add the square ofthe middle term;the square root of the same, less the middle term, being divided by twice thesquareisthevalue. Thiswouldtranslateto4ac+b2 b2aas the solution to the equationax2+bx= can use the Quadratic Formula to solve any Quadratic , thisis shown in the fol-lowing + 3x+ 2 = 0a= 1, b= 3, c= 2,use Quadratic formulax= 3 32 4(1)(2) 2(1)Evaluate exponent and multiplicationx= 3 9 8 2 Evaluate subtraction under rootx= 3 1 2 Evaluate rootx= 3 12 Evaluate to get two answersx= 22or 42 Simplify fractionsx= 1or 2 Our SolutionAs we are solving using the Quadratic Formula , it is important to remember theequation must fist be equal to +11 First set equal to zero 30x 11 30x 11 Subtract 30xand 11 from both sides25x2 30x 11= 0a=25, b= 30, c= 11,use Quadratic formulax=30 ( 30)2 4(25)( 11) 2(25)Evaluate exponent and multiplication2x=30 900+1100 50 Evaluate addition inside rootx=30 2000 50 Simplify rootx=30 205 50 Reduce fraction by dividing each term by 10x=3 2 5 5 Our SolutionExample + 4x+ 8 = 2x2+ 6x 5 First set equation equal to zero 2x2 6x+ 5 2x2 6x+ 5 Subtract2x2and6xand add5x2 2x+13= 0a= 1, b= 2, c=13,use Quadratic formulax=2 ( 2)2 4(1)(13) 2(1)Evaluate exponent and multiplicationx=2 4 52 2 Evaluate subtraction inside rootx=2 48 2 Simplify rootx=2 4i3 2 Reduce fraction by dividing each term by2x= 1 2i3 Our SolutionWhen we use the Quadratic Formula we don t necessarily get two unique can end up with only one solution if the square root simplifies to 12x+ 9 = 0a= 4, b= 12, c= 9,use Quadratic formulax=12 ( 12)2 4(4)(9) 2(4)

3 Evaluate exponents and multiplicationx=12 144 144 8 Evaluate subtraction inside rootx=12 0 8 Evaluate rootx=12 08 Evaluate x=128 Reduce fractionx=32 Our Solution3If a term is missing from the Quadratic , we can still solve with the Quadratic for-mula, we simply use zero for that term. The order is important, so if the termwithxis missing, we haveb= 0, if the constant term is missing, we havec= + 7 = 0a= 3, b= 0(missing term), c= 7x= 0 02 4(3)(7) 2(3)Evaluate exponnets and multiplication,zeros not neededx= 84 6 Simplify rootx= 2i21 6 Reduce,dividing by2x= i21 3 Our SolutionWe have covered three different methods to use to solve a Quadratic : factoring,complete the square, and the Quadratic Formula . It is important to be familiarwith all three as each has its advantage to solving Quadratics . The following tablewalks through a suggested process to decide which method would be best to usefor solving a If it can easily factor, solve by factoringx2 5x+ 6 = 0(x 2)(x 3) = 0x= 2orx= 32. Ifa= 1andbis even, complete the squarex2+ 2x= 4(12 2)2= 12= 1x2+ 2x+ 1 = 5(x+ 1)2= 5x+ 1 = 5 x= 1 5 3.

4 Otherwise, solve by the Quadratic formulax2 3x+ 4 = 0x=3 ( 3)2 4(1)(4) 2(1)x=3 i7 2 The above table is mearly a suggestion for deciding how to solve a completing the square and Quadratic Formula will always work to solveany Quadratic . Factoring only woks if the equation can be and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License. ( ) Practice - Quadratic FormulaSolve each equation with the Quadratic )4a2+ 6 = 03)2x2 8x 2 = 05)2m2 3 = 07)3r2 2r 1 = 09)4n2 36= 011)v2 4v 5 = 813)2a2+ 3a+14= 615)3k2+ 3k 4 = 717)7x2+ 3x 16= 219)2p2+ 6p 16= 421)3n2+ 3n= 323)2x2= 7x+4925)5x2= 7x+ 727)8n2= 3n 829)2x2+ 5x= 331)4a2 64= 033)4p2+ 5p 36= 3p235) 5n2 3n 52= 2 7n237)7r2 12= 3r39)2n2 9 = 42)3k2+ 2 = 04)6n2 1 = 06)5p2+ 2p+ 6 = 08)2x2 2x 15= 010)3b2+ 6 = 012)2x2+ 4x+12= 814)6n2 3n+ 3 = 416)4x2 14= 218)4n2+ 5n= 720)m2+ 4m 48= 322)3b2 3 = 8b24)3r2+ 4 = 6r26)6a2= 5a+1328)6v2= 4 + 6v30)x2= 832)2k2+ 6k 16= 2k34) 12x2+x+ 7 = 5x2+ 5x36)7m2 6m+ 6 = m38)3x2 3 =x240)6b2=b2+ 7 bBeginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License.

5 ( ) - Quadratic Formula1)i6 2, i6 22)i6 3, i6 33)2 + 5 ,2 5 4)6 6, 6 65)6 2, 6 26) 1 +i29 5, 1 i29 57)1, 138)1 +31 2,1 31 29)3, 310)i2 , i2 11)3,112) 1 +i, 1 i13) 3 +i55 4, 3 i55 414) 3 +i159 12, 3 i159 1215) 3 +141 6, 3 141 616)3 , 3 17) 3 +401 14, 3 401 1418) 5 +137 8, 5 137 819)2, 520)5, 921) 1 +i3 2, 1 i3 222)3, 1323)72, 724) 3 +i3 3, 3 i3 325)7 + 321 10,7 321 1026) 5 +337 12, 5 337 1227) 3 +i247 16, 3 i247 1628)3 +33 6,3 33 629) 1, 3230)2 2 , 2 2 31)4, 432)2, 433)4, 934)2 + 3i5 7,2 3i5 735)6, 9236)5 +i143 14,5 i143 1437) 3 +345 14, 3 345 1438)6 2, 6 239)26 2, 26 240) 1 +141 10, 1 141 10 Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License. ( )6