9.4 Quadratics - Quadratic Formula - CCfaculty.org

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.

2 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. 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)

3 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)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.

4 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. 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 .

5 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.

6 ( ) - 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