Chapter 9: Quadratic Equations

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1 Chpter 9: Qudrtic Equtions QUADRATIC EQUATIONS DEFINITION + + c = 0,, c re constnts (generlly integers) ROOTS Synonyms: Solutions or Zeros Cn hve 0, 1, or rel roots Consider the grph of qudrtic equtions. The qudrtic eqution looks like + + c = 0, ut if we tke the qudrtic epression on the left nd set it equl to y, we will hve function: y = + + c When we grph y vs., we find tht we get curve clled prolerror! Bookmrk not defined.. The specific vlues of,, nd c control where the curve is reltive to the origin (left, right, up, or down), nd how rpidly it spreds out. Also, if is negtive then the prol will e upside-down. Wht does this hve to do with finding the solutions to our originl qudrtic eqution? Well, whenever y = 0 then the eqution y = + + c is the sme s our originl eqution. Grphiclly, y is zero whenever the curve crosses the -is. Thus, the solutions to the originl qudrtic eqution ( + + c = 0) re the vlues of where the function (y = + + c) crosses the -is. From the figures elow, you cn see tht it cn cross the -is once, twice, or not t ll. 113

2 No Solutions One Solution Two Solutions Actully, if you hve grphing clcultor this technique cn e used to find solutions to ny eqution, not just qudrtics. All you need to do is 1. Move ll the terms to one side, so tht it is equl to zero. Set the resulting epression equl to y (in plce of zero) 3. Enter the function into your clcultor nd grph it 4. Look for plces where the grph crosses the -is Your grphing clcultor most likely hs function tht will utomticlly find these intercepts nd give you the -vlues with gret precision. Of course, no mtter how mny deciml plces you hve it is still just n pproimtion of the ect solution. In rel life, though, close pproimtion is often good enough. SOLVING BY SQUARE ROOTS NO FIRST-DEGREE TERM If the qudrtic hs no liner, or first-degree term (i.e. = 0), then it cn e solved y isolting the nd tking squre roots of oth sides: + c = 0 = c = c = ± c 114

3 You need oth the positive nd negtive roots ecuse = either positive or negtive., so could e This is only going to give rel solution if either or c is negtive (ut not oth) SOLVING BY FACTORING Solving qudrtic (or ny kind of eqution) y fctoring it mkes use of principle known s the zero-product rule. Zero Product Rule Emple: If = 0 then either = 0 or = 0 (or oth). In other words, if the product of two things is zero then one of those two things must e zero, ecuse the only wy to multiply something nd get zero is to multiply it y zero. Thus, if you cn fctor n epression tht is equl to zero, then you cn set ech fctor equl to zero nd solve it for the unknown. The epression must e set equl to zero to use this principle You cn lwys mke ny eqution equl to zero y moving ll the terms to one side. Given: = 6 Move ll terms to one side: 6 = 0 Fctor: ( 3)( + ) = 0 Set ech fctor equl to zero nd solve: ( 3) = 0 OR ( + ) = 0 Solutions: = 3 OR = 115

4 NO CONSTANT TERM If qudrtic eqution hs no constnt term (i.e. c = 0) then it cn esily e solved y fctoring out the common from the remining two terms: + = 0 ( + ) = 0 Then, using the zero-product rule, you set ech fctor equl to zero nd solve to get the two solutions: = 0 or + = 0 = 0 or = / WARNING: Do not divide out the common fctor of or you will lose the = 0 solution. Keep ll the fctors nd use the zero-product rule to get the solutions. TRINOMIALS When qudrtic hs ll three terms, you cn still solve it with the zero-product rule if you re le to fctor the trinomil. Rememer, not ll trinomil qudrtics cn e fctored with integer constnts If it cn e fctored, then it cn e written s product of two inomils. The zeroproduct rule cn then e used to set ech of these fctors equl to zero, resulting in two equtions tht re oth simple liner equtions tht cn e solved for. See the ove emple for the zero-product rule to see how this works. A more thorough discussion of fctoring trinomils my e found in the chpter on polynomils, ut here is quick review: TIPS FOR FACTORING TRINOMIALS 1. Cler frctions (y multiplying through y the common denomintor). Remove common fctors if possile 3. If the coefficient of the term is 1, then + + c = ( + n)( + m), where n nd m i. Multiply to give c ii. Add to give 4. If the coefficient of the term is not 1, then use either. Guess-nd Check i. List the fctors of the coefficient of the term 116

5 ii. List the fctors of the constnt term iii. Test ll the possile inomils you cn mke from these fctors. Fctoring y Grouping i. Find the product c ii. Find two fctors of c tht dd to give iii. Split the middle term into the sum of two terms, using these two fctors iv. Group the terms into pirs COMPLETING THE SQUARE The technique of completing the squre is presented here primrily to justify the qudrtic formul, which will e presented net. However, the technique does hve pplictions esides eing used to derive the qudrtic formul. In nlytic geometry, for emple, completing the squre is used to put the equtions of conic sections into stndrd form. Before considering the technique of completing the squre, we must define perfect squre trinomil. Perfect Squre Trinomil Wht hppens when you squre inomil? ( + ) = + + Note tht the coefficient of the middle term () is twice the squre root of the constnt term ( ) Thus the constnt term is the squre of hlf the coefficient of Importnt: These oservtions only hold true if the coefficient of is 1. This mens tht ny trinomil tht stisfies this condition is perfect squre. For emple, is perfect squre, ecuse hlf the coefficient of (which in this cse is 4) hppens to e the squre root of the constnt term (16). Tht mens tht = ( + 4) Multiply out the inomil ( + 4) times itself nd you will see tht this works. 117

6 The technique of completing the squre is to tke trinomil tht is not perfect squre, nd mke it into one y inserting the correct constnt term (which is the squre of hlf the coefficient of ). Of course, inserting new constnt term hs to e done in n lgericlly legl mnner, which mens tht the sme thing needs to e done to oth sides of the eqution. This is est demonstrted with n emple. Emple: Given Eqution: + 6 = 0 Move originl constnt to other side: + 6= Add new constnt to oth sides (the squre of hlf the coefficient of ): = + 9 Write left side s perfect squre: ( + 3) = 11 Squre root oth sides (rememer to use plus-or-minus): + 3 = ± 11 Solve for : = 3 ± 11 Notes Finds ll rel roots. Fctoring cn only find integer or rtionl roots. When you write it s inomil squred, the constnt in the inomil will e hlf of the coefficient of. IF THE COEFFICIENT OF X IS NOT 1 First divide through y the coefficient, then proceed with completing the squre. Emple: Given Eqution: + 3 = 0 1 Divide through y coefficient of : + 3 = 0 (in this cse ) = 0 Move constnt to other side: Add new constnt term: (the squre of hlf the coefficient of, in this cse 9/16): Write s inomil squred: (the constnt in the inomil is hlf the coefficient of ) Squre root oth sides: (rememer to use plus-or-minus) Solve for : ( ) 3 + = = ( + ) = = ± ± 5 = 4 118

7 Thus = ½ or = THE QUADRATIC FORMULA The solutions to qudrtic eqution cn e found directly from the qudrtic formul. The eqution + + c = 0 hs solutions = ± 4c The dvntge of using the formul is tht it lwys works. The disdvntge is tht it cn e more time-consuming thn some of the methods previously discussed. As generl rule you should look t qudrtic nd see if it cn e solved y tking squre roots; if not, then if it cn e esily fctored; nd finlly use the qudrtic formul if there is no esier wy. Notice the plus-or-minus symol (±) in the formul. This is how you get the two different solutions one using the plus sign, nd one with the minus. Mke sure the eqution is written in stndrd form efore reding off,, nd c. Most importntly, mke sure the qudrtic epression is equl to zero. THE DISCRIMINANT The formul requires you to tke the squre root of the epression 4c, which is clled the discriminnt ecuse it determines the nture of the solutions. For emple, you cn t tke the squre root of negtive numer, so if the discriminnt is negtive then there re no solutions. If 4c > 0 If 4c = 0 If 4c < 0 There re two distinct rel roots There is one rel root There re no rel roots 119

8 DERIVING THE QUADRATIC FORMULA The qudrtic formul cn e derived y using the technique of completing the squre on the generl qudrtic formul: Given: + + c = 0 c Divide through y : + + = 0 Move the constnt term to the right side: Add the squre of one-hlf the coefficient of to oth sides: Fctor the left side (which is now perfect squre), nd rerrnge the right side: + + = c + = 4 + = c c + Get the right side over common 4c denomintor: + = 4 Tke the squre root of oth sides (rememering to use plus-or-minus): + = ± 4c Solve for : ± = 4c 10

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