QUA DR ATIC EQUATION

J-Mthemtics. INTRODUCTION : QUA DR ATIC QUATION The lgebric epression of the form + b + c, 0 is clled qudrtic epression, becuse the highest order term in it is of second degree. Qudrtic eqution mens, + b + c = 0. In generl whenever one ss zeroes of the epression +b + c, it implies roots of the eqution + b + c = 0, unless specified otherwise. A qudrtic eqution hs ectl two roots which m be rel (equl or unequl) or imginr.. SOLUTION OF QUA DR ATIC QUATION & RL ATION BT WN ROOTS & CO- FFICINTS : ( ) The generl form of qudrtic eqution is + b + c = 0, 0. The roots cn be found in following mnner : b c = 0 b b c b c b 0 b b c This epression cn be directl used to find the two roots of qudrtic eqution. ( b ) The epression b c D is clled the discriminnt of the qudrtic eqution. ( c ) If & re the roots of the qudrtic eqution + b + c = 0, then : (i) = b/ (ii) c / (iii) D / ( d ) A qudrtic eqution whose roots re & is ( ) ( ) = 0 i.e. ( ) + = 0 i.e. (sum of roots) + product of roots = 0. Illustrtion : If re the roots of qudrtic eqution 3 + 5 = 0, then the eqution whose roots re ( 3 + 7) nd ( 3 + 7) is - (A) + + = 0 (B) + = 0 (C) = 0 (D) + + 3 = 0 Since re the roots of eqution 3 + 5 = 0 So 3 + 5 = 0 3 + 5 = 0 3 = 5 3 = 5 Putting in ( 3 + 7) & ( 3 + 7) 5 + 7, 5 + 7 nd re the roots. The required eqution is 30...(i) + = 0. Ans. (B) Illustrtion : If nd re the roots of + b + c = 0, find the vlue of ( + b) + ( + b). b We know tht + = & = c ( + b) + ( + b) = ( b) ( b) = b b b b ( ) b( ) b = ( b b b ) ( b( ) b ) ( + cn lws be written s ( + ) ) NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY

J-Mthemtics ( ) b( ) b = ( b( ) b ) Alternt ivel : Tke b = (+ ) b c b b b = c b b b ( + b) + ( + b) = ( ) ( ) b c = c = b c b c = c c. Do ourself - : ( i ) Find the roots of following equtions : () + 3 + = 0 (b) 8 + 6 = 0 (c) = 0 ( i i ) Find the roots of the eqution ( + ) ( + ) = 0, where 0. (iii) 6 Solve : ( i v ) If the roots of + 5k = (5k + ) differ b unit, then find the vlues of k. 3. NATUR OF ROOTS : ( ) Consider the qudrtic eqution + b + c = 0 where, b, c R & 0 then ; NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY (i) (ii) (iii) (iv) b D D > 0 roots re rel & distinct (unequl). D = 0 roots re rel & coincident (equl) D < 0 roots re imginr. If p + i q is one root of qudrtic eqution, then the other root must be the conjugte p i q & vice vers. (p, q R & i = ). ( b ) Consider the qudrtic eqution + b + c = 0 where, b, c Q & 0 then ; (i) Illustrtion 3 : If D is perfect squre, then roots re rtionl. (ii) If = p + q is one root in this cse, ( where p is rtionl & q is surd) then other root will be p q. If the coefficient of the qudrtic eqution re rtionl & the coefficient of is, then find the eqution one of whose roots is tn 8. We know tht tn 8 Irrtionl roots lws occur in conjugtionl pirs. Hence if one root is ( + ), the other root will be ( ). qution is ( ( + )) ( ( )) =0 + = 0 3

J-Mthemtics Illustrtion : Do ourself - : Find ll the integrl vlues of for which the qudrtic eqution ( )( 0) + = 0 hs integrl roots. Here the eqution is ( + 0) + 0 + = 0. Since integrl roots will lws be rtionl it mens D should be perfect squre. From (i) D = 0 + 96. D = ( 0) = ( 0) D If D is perfect squre it mens we wnt difference of two perfect squre s which is possible onl when ( 0) = nd D = 0. ( 0) = ± =, 8 A n s. ( i ) If 3 is root of the eqution + b + c = 0, where b, c Q, find b, c. ( i i ) For the following equtions, find the nture of the roots (rel & distinct, rel & coincident or imginr). () 6 + 0 = 0 (iii) (b) (7 3 ) 6( 3 ) 0 (c) + 8 + 9 = 0 If, m re rel nd m, then show tht the roots of ( m) 5( + m) ( m) = 0 re rel nd unequl.. ROOTS UNDR PARTICUL AR CASS : Let the qudrtic eqution + b + c = 0 hs rel roots nd ( ) If b = 0 roots re equl in mgnitude but opposite in sign ( b ) If c = 0 one root is zero other is b/ ( c ) If = c roots re reciprocl to ech other ( d ) If 0 c 0 0 c 0 0, b 0,c 0 ( e ) If 0, b 0,c 0 roots re of opposite signs both roots re negtive. 0, b 0,c 0 ( f ) If 0, b 0,c 0 both roots re positive. ( g ) If sign of = sign of b sign of c Greter root in mgnitude is negtive. ( h ) If sign of b = sign of c sign of Greter root in mgnitude is positive. ( i ) If + b + c = 0 one root is nd second root is c/ or ( b )/. Illustrtion 5 : If eqution is - b k c k hs roots equl in mgnitude & opposite in sign, then the vlue of k (A) b (B) b b b (C) b (D) b Let the roots re. given eqution is ( b)(k + ) = (k )( c) {Considering, c/ & k } (k + ) b(k + ) = (k ) c(k ) (k + ) b(k + ) (k ) + c(k ) = 0 Now sum of roots = 0 ( = 0) b(k + ) + (k ) = 0 k = b b 3 Ans. (B) NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY

J-Mthemtics *Illustrtion 6 : If roots of the eqution ( b) + (c ) + (b c) = 0 re equl, then, b, c re in (A) A.P. (B) H.P. (C) G.P. (D) none of these ( b) + (c ) + (b c) = 0 Do ourself - 3 : As roots re equl so B AC = 0 (c ) ( b)(b c) = 0 (c ) b + b + c bc = 0 (c ) + c b(c + ) + b = 0 (c + ). (b)(c + ) + (b) = 0 [c + b] = 0 c + b = 0 c + = b Hence, b, c re in A. P. Alternt ive met ho d : Sum of the coefficients = 0 Hence one root is nd other root is b c. b Given tht both roots re equl, so = b c b = b c b = + c b Hence, b, c re in A.P. ( i ) Consider ƒ () = + b + c. () Find c if = 0 is root of ƒ () = 0. (b) 5. IDNTITY : Find c if, re roots of ƒ () = 0. (c) Comment on sign of b & c, if 0 &, where re roots of ƒ () = 0. Ans. (A) An eqution which is true for ever vlue of the vrible within the domin is clled n identit, for emple : 5 ( 3) =5 5, ( + b) = + b + b for ll, b R. Note : A qudrtic eqution cnnot hve three or more roots & if it hs, it becomes n identit. If + b + c = 0 is n identit = b = c = 0 NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY Illustrtion 7 : If the eqution ( 5 + 6) + ( 3 + ) + ( ) = 0 hs more thn two roots, then find the vlue of? As the eqution hs more thn two roots so it becomes n identit. Hence 5 + 6 = 0 =, 3 nd 3 + = 0 =, nd = 0 =, So = Ans. = 6. COMMON ROOTS OF T WO QUA DR ATIC QUATIONS : ( ) Onl one common root. Let be the common root of + b + c = 0 & ' + b' + c' = 0 then + b + c = 0 & ' + b' + c' = 0. B Crmer s Rule Therefore, c ' c ' bc ' b 'c b ' 'b 'c c ' So the condition for common root is (c' - c') = (b' - 'b) (bc'- b'c). 33 bc ' b 'c 'c c ' b ' 'b

J-Mthemtics ( b ) If both roots re sme then b c. ' b ' c ' Illustrtion 8 : Find p nd q such tht p + 5 + = 0 nd 3 + 0 +q = 0 hve both roots in common. = p, b = 5, c = = 3, b = 0, c = q We know tht : b c b c p 5 p = 3 3 0 q ; q = *Illustrtion 9 : The equtions 5 + + 3 = 0 nd, b, c re the sides of the ABC. Then find C. + b + c = 0 (,b,c R) hve common root, where (A) 5 (B) 60 (C) 90 (D) 30 As we cn see discriminnt of the eqution 5 + + 3 = 0 is negtive so roots of the eqution re imginr. We know tht imginr roots lws occurs in pir. So this eqution cn not hve single common roots with n other eqution hving rel coefficients. So both roots re common of the given equtions. b c Hence (let) 5 3 then = 5, b =, c = 3 Now cosc = b c 5 69 b (5 )( ) 0 C = 90 Ans. (C) Do ourself - : ( i ) If + b + c = 0 & + 9 + 0 = 0 hve both roots, find b & c. ( i i ) If 7 + 0 = 0 & 5 + c = 0 hve common root, find c. (iii) Show tht + ( ) = 0 nd 3 + = 0 hve ectl one common root for ll R. 7. RMAINDR THORM : If we divide polnomil f() b ( ) the reminder obtined is f(). If f() is 0 then ( ) is fctor of f(). Consider f() = 3 9 + 3 5 f() = 0 ( ) is fctor of f(). f() = ( )( 7 + 9) + 3. Hence f() = 3 is reminder when f() is divided b ( ). 8. SOLUTION OF R ATIONAL INQUALITIS : Let f() be n epression in where f() & g() re polnomils in. Now, if it is given tht g() > 0 (or < 0 or > 0 or < 0), this clls for ll the vlues of for which stisfies the constrint. This solution set cn be found b following steps : Step I : Fctorize f() & g() nd generte the form : ( ) ( )...( ) ( b ) ( b )...( b ) n n n k k m m m p p where n n...n k, m,m...m p re nturl numbers nd,... k, b,b...b p re rel numbers. Clerl, here,... k re roots of f()=0 & b,b,...b p re roots of g() = 0. 3 NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY

J-Mthemtics Step II : Here vnishes (becomes zero) for,,... k. These points re mrked on the number line with blck dot. The re solution of =0. f() If g()=0, ttins n undefined form, hence b g(),b...b k re ecluded from the solution. These points re mrked with white dots. Step-III : Step-IV : Step-V : Note : e.g. 3 5 ( ) 3 6 f() 3 7 6 0 3 7 Check the vlue of for n rel number greter thn the right most mrked number on the number line. If it is positive, then is positive for ll the rel numbers greter thn the right most mrked number nd vice vers. If the eponent of fctor is odd, then the point is clled simple point nd if the eponent of fctor is even, then the point is clled double point 3 5 ( ) 3 6 7 3 Here,3, 6 nd 7 re simple points nd & 0 re double points. From right to left, beginning bove the number line (if is positive in step 3 otherwise from below the line), wv curve should be drwn which psses through ll the mrked points so tht when pssing through simple point, the curve intersects the number line nd when pssing through double point, the curve remins on the sme side of number line. f() 3 5 ( ) ( ) ( 3) ( 6) 3 ( 7) 6 0 3 7 As eponents of ( + ) nd re even, the curve does not cross the number line. This method is clled wv curve method. The intervls where the curve is bove number line, will be positive nd the intervls where the curve is below the number line, will be negtive. The pproprite intervls re chosen in ccordnce with the sign of inequlit & their union represents the solution of inequlit. NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY (i) (ii) (iii) (iv) (v) (vi) Points where denomintor is zero will never be included in the nswer. If ou re sked to find the intervls where f() is non-negtive or non-positive then mke the intervls closed corresponding to the roots of the numertor nd let it remin open corresponding to the roots of denomintor. Normll we cnnot cross-multipl in inequlities. But we cross multipl if we re sure tht quntit in denomintor is lws positive. Normll we cnnot squre in inequlities. But we cn squre if we re sure tht both sides re non negtive. We cn multipl both sides with negtive number b chnging the sign of inequlit. We cn dd or subtrct equl quntit to both sides of inequlities without chnging the sign of inequlit. Illustrtion 0 : Find such tht 3 7 + 6 < 0 D = 9 7 < 0 As D < 0, 3 7 + 6 will lws be positive. Hence. 35

J-Mthemtics Illustrtion : ( 6) ( + 6) > 0 ( 3) (+) () (+6) > 0 Consider = ( 3)(+)(+6), = 0 = 0, 3,, 6 (ll re simple points) 6 0 3 For > 3 = ( 3) (+) (+6) +ve +ve +ve +ve = positive 6 0 3 Hence for ( 3) (+) (+6) > 0 (, 6] [,0] [3,) 3 5 ( ) ( ) ( 3) ( 6) Illustrtion : Let f() =. Solve the following inequlit 3 ( 7) (i) f() > 0 (ii) f() 0 (iii) f() < 0 (iv) f() 0 We mrk on the number line zeros of the function :,, 3 nd 6 (with blck circles) nd the points of discontinuit 0 nd 7 (with white circles), isolte the double points : nd 0 nd drw the wv curve : 0 6 3 7 From grph, we get (i) (, 6) (, 3) (7, ) (ii) (, 6] { } [, 3] (7, ) (iii) ( 6, ) (, 0) (0, ) (3, 7) (iv) [ 6, 0) (0, ] [3, 7) Do ourself - 5 : ( i ) Find rnge of such tht () ( )( + 3) 0 (b) (c) 3 0 7 7 (e) 3 9. QUA DR ATIC XPRSSION AND IT' S GR APHS : 36 (d) ( )( 3)( )( ) 0 ( 6)( 9)( 9) (f) + 3 5 Consider the qudrtic epression, = + b + c, 0 &, b, c R then ; ( ) The grph between, is lws prbol. If > 0 then the shpe of the prbol is concve upwrds & if < 0 then the shpe of the prbol is concve downwrds. ( b ) The grph of = + b + c cn be divided in 6 brod ctegories which re s follows : (Let the rel roots of qudrtic eqution + b + c = 0 be & where < ). NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY

J-Mthemtics Fig. Fig. F i g. 3 > 0 D > 0 > 0 D = 0 > 0 D < 0 O b, D O b, D O b, D Roots re rel & distinct Roots re coincident Roots re comple conjugte + b + c > 0 (, ) (, ) + b + c > 0 R {} + b + c > 0 R + b + c < 0 (, ) + b + c = 0 for = = b, D O Fig. Fig. 5 F i g. 6 < 0 D > 0 b, D O < 0 D = 0 O b, D < 0 D < 0 Root s re rel & dist inct Roots re coincident Roots re comple conjugte + b + c > 0 (, ) +b + c < 0 R {} + b + c < 0 R + b + c < 0 (, ) (, ) + b + c = 0 for = Importnt Note : (i) The qudrtic epression + b + c > 0 for ech R > 0, D < 0 & vice-vers (Fig. 3) (ii) The qudrtic epression + b + c < 0 for ech R < 0, D < 0 & vice-vers (Fig. 6) NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY 0. MA XIMUM & MINIMUM VALUS OF QUADRATIC XPRSSIONS : = + b + c : We know tht = b (b c) + b + c tkes following form : =, b D which is prbol. verte =, b D When > 0, will tke minimum vlue t verte; = ; min When < 0, will tke mimum vlue t verte; = b D ;. m If qudrtic epression +b +c is perfect squre, then > 0 nd D = 0 *Illustrtion3 : If f() is qudrtic epression such tht f() > 0 R, nd if g() = f() + f'() + f''(), then prove tht g() > 0 R. Let f() = + b + c Given tht f() > 0 so > 0, b c < 0 Now g() = + b + c + + b + = + (b + ) + (b + c + ) For this qudrtic epression > 0 nd discriminnt D = (b + ) (b + c + ) = b + + b b c 8 = b c < 0 So g() > 0 R. 37

J-Mthemtics Illustrtion : The vlue of the epression b c will be positive for ll rel if - (A) b c 0 (B) b c 0 (C) c b (D) b c As > 0, so this epression will be positive if D < 0 so b c < 0 b < c A ns. ( D ) Illustrtion 5 : The minimum vlue of the epression + + is - (A) / (B) / (C) 3/ (D) Since = > 0 therefore its minimum vlue is = 38 ()() () 6 3 *Illustrtion6 : If = 3, then find the rnge of when : (i) R (ii) [0,3] (iii) [,0] We know tht minimum vlue of will occur t Ans. (C) () 6 6 b ( ) O D ( 3 ) min = = (, ) (i) R; [,) A n s. (ii) [0, 3] f(0) = 3, f() =,f(3) = 0 f(3) > f(0) will tke ll the vlues from minimum to f(3). [, 0] A n s. 5 (iii) [, 0] This intervl does not contin the minimum vlue of for R. O will tke vlues from f(0) to f( ) f(0) = 3 (0, 3) f( ) = 5 [ 3, 5] A n s. Illustrtion 7 : If + b + 0 = 0 does not hve rel & distinct roots, find the minimum vlue of 5 b. ither f() 0 R or f() 0 R f(0) = 0 > 0 f() 0 R f( 5) = 5 5b + 0 5 b A n s. Do ourself - 6 ( i ) Find the minimum vlue of : () = + + (b) = 6 + 5 ( i i ) For following grphs of = + b + c with,b,c R, comment on the sign of : (i) (ii) b (iii) c (iv) D (v) + (vi) (iii) () () (3) = =0 Given the roots of eqution + b + c = 0 re rel & distinct, where,b,c R +, then the verte of the grph will lie in which qudrnt. *(iv) Find the rnge of '' for which : () + 3 + > 0 R (b) + < 0 R NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY

J-Mthemtics. INQUALITIS IN VOLVING MODULUS FUNCTION : Proper t ie s of mo dulus funct ion : (i) (ii) or, where is positive. [, ], where is positive (iii) > > (iv) b b b (v) + = + 0 (vi) = + 0 Illustrtion 8 : If stisfies + + 3 6, then (A) 0 (B) or (C) 0 or Cse I :, then Cse II : (D) none of these + + 3 6 0 Hence < 0 <, then + + 3 6 But < < No solution. Cse III : < 3, then + + 3 6 6 But < < 3 No solution. Cse IV : > 3, then + + 3 6 Hence >...(i)...(ii)...(iii)...(iv) From (i), (ii), (iii) nd (iv) the given inequlit holds for 0 or. NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY Illustrtion 9 : Solve for : () +. (b) () + + 6 3 [, 3] (b) Cse : Given ineqution will be sttisfied for ll such tht 0 (, ] {}...(i) (Note : {} is not in domin of RHS) Cse : 0 (, ) (, )...(ii) Given ineqution becomes on solving we get 39 or on solving we get, / 5 (, ) (, 0] (, 5 / ] tking intersection with (ii) we get (, )...(iii) tking intersection with (ii) we get Hence, solution of the originl ineqution : (,) {} (tking union of (i) & (iii))

J-Mthemtics Illustrtion 0 : The eqution + is lws true for belongs to (A) {0} (B) (, ) (C) (, ) (D) (, ) + =. IRR ATIONAL INQUALITIS : = is true onl if. 0 {0} (, ). Ans (A, B) Illustrtion : Solve for, if 3 3 0 ( )( ) 0 0 ( ) 0 ( 3 ) ( ) 0 or or 3 0 ( )( ) 0 0 0 Hence, solution set of the originl ineqution is R (,] Do ourself - 7 : ( i ) Solve for if ( i i ) Solve for if ( ) 3. LOGARITHMIC INQUALITIS : Poi nt s to remember : < if > ( i ) log < log > if 0 < < p p ( i i ) If >, then () log < p 0 < < (b) log > p > p p (iii) If 0 < <, then () log < p > (b) log > p 0 < < Illustrtion : Solve for : () log 0.5 ( 5 + 6) > (b) log /3 (log ( 5)) > 0 () log 0.5 ( 5 + 6) > 0 < 5 + 6 < (0.5) 0 < 5 + 6 < 5 6 0 [,) (3, ] 5 6 Hence, solution set of originl ineqution : [,) (3,] 0 NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY

J-Mthemtics (b) log /3 (log ( 5)) > 0 0 < log ( 5) < 0 log ( 5) 5 > log ( 5) < 0 < 5 < < ( 5) < 6 < < 9 3, 6 6, 3 Hence, solution set of originl ineqution : 3, 6 6, 3 Illustrtion 3 : Solve for : log log. Let log = t t t t 0 t t t 0 (t )(t ) 0 t (t ) t (, ] (,] or log (, ] (,] or 0, (,] Illustrtion : Solve the ineqution : log log ( 3) 3 3 This ineqution is equivlent to the collection of the sstems 3 or 0 3 0 3 3 0 ( 3)( ) 0 & 0 or 3 ( 3)( ) 0 3 & 0 3 or or 3 3 or 3 Hence, solution of the originl ineqution is 3, (, 0) (0, 3) NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY. XPONNTIAL INQUATIONS : If f() > b (0, ) O Illustrtion 5 : Solve for : =, > f() log b when f() log b when 0 (0, ) O =, 0 < <

J-Mthemtics We hve + > /. Since the bse >, we hve + > retined). Now 0 0 (the sign of the inequlit is ( ) 0 (0, ) ( ) Illustrtion 6 : Solve for : (.5) (0.6) We hve ( ) 5 6 5 or ( ) 5 5 Since the bse 5 Now, R.H.S. is positive 0, the inequlit is equivlent to the inequlit > ( ) 5 5 0 > 5...(i) we hve 5 both sides re positive, so squring both sides ( 5) ( 5) or 0 6 6 or 6 + 5 > 0 or ( 5) ( ) > 0 (, ) (5, )...(ii) intersection (i) & (ii) gives (5, ) Do ourself- 8 : ( i ) Solve for : () log 0.3 ( 6 + 8) > log 0.3 (9) (b) log7 0 ( i i ) Solve for : 3 5. MA XIMUM & MINIMUM VALUS OF R ATIONAL ALGBR AIC XPRSSIONS : = b c b c, b c, b b c Sometime we hve to find rnge of epression of form Step : qute the given epression to i.e., b c b b c b c b c b c Step : B cross multipling nd simplifing, obtin qudrtic eqution in. ( ) + (b b ) + (c c ) = 0 Step 3 : Put Discriminnt 0 nd solve the inequlit for possible set of vlues of. :. The following procedure is used : NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY

J-Mthemtics Illustrtion 7 : For R, find the set of vlues ttinble b 3. 3 Illustrtion 8 3 Let 3 ( ) + 3( + ) + ( ) = 0 Cse- I : For bove eqution is qudrtic eqution. So for R, D 0 9( + ) 6( ) 0 7 50 + 7 0 (7 )( 7) 0 Cse II : when = 3 3 + 3 + = 3+ = 0 Hence = for rel vlue of. so rnge of is,7 7,7 {} 7 :Find the vlues of for which the epression of. Let = 3 3 ( + ) + 3( ) ( + ) = 0 If R, D > 0 3 3 ssumes ll rel vlues for rel vlues 9( ) + ( + )( + ) 0 (9 + 6) + ( + 6) + (9 + 6) 0 for ll R, (9 + 6) > 0 & D 0 ( + 6) (9 + 6)(9 + 6) 0 ( 8 + 7)( + 8 + 6) 0 8 + 7 0 7 NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY 9 + 6 > 0 & 7 Tking intersection, [, 7] Now, checking the boundr vlues of For = 3 ( )( ) 3 ( )( ) = is not possible. if = 7 7 3 3 7 (7 )( ) (7 )( ) So will ssume ll rel vlues for some rel vlues of. So (,7) 3

J-Mthemtics Do ourself - 9 : 8 ( i ) Prove tht the epression cnnot hve vlues between nd, in its domin. ( i i ) Find the rnge of, where is rel 7 6. LOCATION OF ROOTS : This rticle dels with n elegnt pproch of solving problems on qudrtic equtions when the roots re locted / specified on the number line with vriet of constrints : Consider the qudrtic eqution + b + c = 0 with > 0 nd let f() = + b + c Tpe- : Both roots of the qudrtic eqution re greter thn specific number (s d). The necessr nd sufficient condition for this re : (i) D 0 ; (ii) ƒ (d) > 0 ; (iii) b > d Note : When both roots of the qudrtic eqution re less thn specific number d thn the necessr nd sufficient condition will be : (i) D 0 ; (ii) ƒ (d) > 0 ; b (iii) < d Tpe- : Both roots lie on either side of fied number s (d). Alterntivel one root is greter thn 'd' nd other root less thn 'd' or 'd' lies between the roots of the given eqution. The necessr nd sufficient condition for this re : f(d) < 0 Note : Considertion of discriminnt is not needed. Tpe- 3 : ctl one root lies in the intervl (d, e). d d d The necessr nd sufficient condition for this re : ƒ (d). ƒ (e) < 0 d e d e Note : The etremes of the intervls found b given conditions give 'd' or 'e' s the root of the eqution. Hence in this cse lso check for end points. Tpe- : When both roots re confined between the number d nd e (d < e). The necessr nd sufficient condition for this re : (i) D 0; (ii) ƒ (d) > 0 ; (iii) ƒ (e) > 0 b (iv) d e Tpe- 5 : One root is greter thn e nd the other roots is less thn d (d < e). d e d e f(d) = 0 f(e) = 0 The necessr nd sufficient condition for this re : f(d) < 0 nd f(e) < 0 Note : If < 0 in the qudrtic eqution + b + c = 0 then we divide the whole eqution b ''. Now ssume b c s f(). This mkes the coefficient of positive nd hence bove cses re pplicble. d d e e NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY

J-Mthemtics NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY Illustrtion 9 : Find the vlues of the prmeter '' for which the roots of the qudrtic eqution + ( ) + + 5 = 0 re (i) rel nd distinct (ii) equl (iii) opposite in sign (iv) equl in mgnitude but opposite in sign (v) positive (vi) negtive (vii) greter thn 3 (viii) smller thn 3 (i) such tht both the roots lie in the intervl (, 3) Let f() = + ( ) + + 5 = A + B + C (s) A =, B = ( ), C = + 5. Also D = B AC = ( ) ( + 5) = ( + )( ) (i) D > 0 (ii) D = 0 (iii) (iv) (v) (vi) (vii) (viii) (i) ( + )( ) > 0 (, )(, ). ( + )( ) = 0 =,. This mens tht 0 lies between the roots of the given eqution. f(0) < 0 nd D > 0 i.e. (, ) (, ) + 5 < 0 < 5 (, 5). This mens tht the sum of the roots is zero ( ) = 0 nd D > 0 i.e. (, ) (, ) = which does not belong to (, )(, ). This implies tht both the roots re greter thn zero B > 0, C 0, D 0. ( ) > 0, + 5 > 0, (, ][, ) A A <, 5 <, (, ][, ) ( 5, ] This implies tht both the roots re less thn zero B < 0, C 0, D 0. ( ) < 0, + 5 > 0, (, ][, ) A A >, > 5, (, ][, ) [, ). In this cse B 3, A.f(3) > 0 nd D 0. ( ) > 3, 7 + 8 > 0 nd (, ][, ) <, > 8/7 nd (, ][, ) Since no vlue of '' cn stisf these conditions simultneousl, there cn be no vlue of for which both the roots will be greter thn 3. In this cse B 3, A.f(3) > 0 nd D 0. >, > 8/7 nd (, ][, ) ( 8/7, ][, ) In this cse B < < 3, A.f() > 0, A.f(3) > 0, D 0. A < ( ) < 3, 3 + > 0, 7 + 8 > 0, (, ][, ) < < 0, > / 3, > 8/7, (, ][, ) 5 8, 7

J-Mthemtics Illustrtion 30 : Find vlue of k for which one root of eqution (k+) + k + k 8 = 0 eceeds & other is less thn. (k+) + k + k 8 < 0 k k 6 < 0 (k 3) (k+) < 0 < k < 3 Tking intersection, k (, 3). Illustrtion 3 : Find ll possible vlues of for which ectl one root of (+) + = 0 lies in intervl (0,3). f(0). f(3) < 0 (9 3( + )+) < 0 ( + 6) < 0 ( 6) > 0 < 0 or > 6 Checking the etremes. If = 0, = 0 = 0, (0, 3) If = 6, 7 + = 0 = 3, But (0, 3) Hence solution set is (,0] (6,) Illustrtion 3 : If is root of the eqution + b + c = 0 nd is root of the eqution + b + c = 0, then prove tht there will be root of the eqution + b + c = 0 ling between nd. Let f() = + b + c f() = + b + c = + b + c = (As is root of + b + c = 0) And f() = + b + c = + b + c + 3 = 3 (As is root of + b + c = 0) Now 3 f(). f() = 0 f() = 0 hs one rel root between nd. Illustrtion 33 : Let, b, c be rel. If + b + c = 0 hs two rel roots nd where < nd >, then show tht + c b 0. Let f() = + b c from grph f( ) < 0 nd f() < 0 + c b 0 nd + c b < 0 Multipling these two, we get c b 0 c b { < c } + c b 0 6 NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY

J-Mthemtics Do ourself - 0 : ( i ) If re roots of 7 + 9 = 0, find their position with respect to following ( ) : () 3 (b) 0 (c) ( i i ) If >, roots of the eqution ( ) + 3 = 0 re - (A) one positive one negtive (B) both negtive (C) both positive (D) both non-rel (iii) Find the set of vlue of for which the roots of the eqution + + 3 = 0 re less thn 3. ( i v ) If re the roots of 3 + = 0, R nd < <, then find the vlues of. ( v ) If re roots of 6 + = 0, R such tht < < nd < < 3, then find the rnge of. 7. GNR AL QUA DR ATIC XPRSSION IN T WO VARIA BLS : f(, ) = + h + b + g + f + c m be resolved into two liner fctors if ; = bc + fgh f bg ch = 0 OR h g h b f 0 g f c Illustrtion 3 : If + + + m 3 hve two liner fctor then m is equl to - (A) 6, (B) 6, (C) 6, (D) 6, Here =, h =, b = 0, g =, f = m/, c = 3 So = 0 0 m / 0 m / 3 Do ourself - : m ( 3 m/) + m/ = 0 m + m + 3 = 0 m m = 0 m =, 6 Ans. (C) ( i ) Find the vlue of k for which the epression + + k + + k = 0 cn be resolved into two liner fctors. NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY 8. THORY OF QUATIONS : Let,, 3,... n re roots of the eqution, ƒ () = 0 n + n + n +... n + n = 0, where 0,,... n re constnts nd 0 0. ƒ () = 0 ( )( )( 3 )... ( n ) 0 n + n +... n + n = 0 ( )( )... ( n ) Compring the coefficients of like powers of, we get i S 0 (s) coefficient of or S coefficient of S = ij ( ) i j n n 0 7

J-Mthemtics S 3 = ijk ijk ( ) 3 3 0 n n S n =... n ( ) = ( ) constnt term n coefficient of n 0 where S k denotes the sum of the product of root tken k t time. Qudrtic eqution : If re the roots of the qudrtic eqution + b + c = 0, then b c nd Cubic eqution : If re roots of cubic eqution 3 + b + c + d = 0, then b c, nd Note : d (i) If is root of the eqution f () = 0, then the polnomil f() is ectl divisible b ( ) or ( ) (ii) is fctor of f() nd conversel. ver eqution of nth degree ( n ) hs ectl n root & if the eqution hs more thn n roots, it is n identit. (iii) If the coefficients of the eqution f () = 0 re ll rel nd i is its root, then i is lso root. i.e. imgi nr root s occur i n conjugte pir s. (iv) If the coefficients in the eqution re ll rtionl & is one of its roots, then is lso root where, Q & is not perfect squre. (v) If there be n two rel numbers & b such tht f() & f(b) re of opposite signs, then f()=0 must hve tlest one rel root between nd b. (vi) ver eqution f() = 0 of degree odd hs tlest one rel root of sign opposite to tht of its lst term. Descrtes rule of signs : The mimum number of positive rel roots of polnomil eqution ƒ () = 0 is the number of chnges of signs in ƒ (). Consider 3 + 6 + 6 = 0 The signs re : + + + As there is onl one chnge of sign, the eqution hs tmost one positive rel root. The mimum number of negtive rel roots of polnomil eqution ƒ () = 0 is the number of chnges of signs in ƒ ( ) Consider ƒ () = + 3 + = 0 ƒ ( ) = 3 + + = 0 3 sign chnges, hence tmost 3 negtive rel roots. Illustrtion 35 : If two roots re equl, find the roots of 3 + 0 3 + 6 = 0. Let roots be nd = 0 = 3 from eqution (i) + ( 5 ) = 3 = /, 3 6 + = 5... (i) + = 3 & = 6 8 0 = 3 + 0 3 = 0 NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY

J-Mthemtics when = = ( 5 ) = 3 when = 3 6 = 3 3 3 3 5 36 6 = = 6 Hence roots of eqution =,, 6 A n s. Illustrtion 36 : If re the roots of 3 p + q r = 0, find : (i) 3 (ii) We know tht = p = q = r (i) 3 3 3 ( ) ( ) ( ) = 3 ( ){( ) 3( )} = 3r + p{p 3q} = 3r + p 3 3pq (ii) ( ) ( ) ( ) = (p ) (p ) (p ) = 3 p( ) 3r p 3pq = p(p q) 3r p 3 + 3pq = pq 3r Illustrtion 37 : If b < c nd, b, c, d R, then prove tht 3 + b + c + d = 0 hs ectl one rel root. Let be the roots of 3 + b + c + d = 0 Then = b + + = c NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY = d + + = () () = 9 b c b c + + < 0, which is not possible if ll re rel. So tlest one root is non-rel, but comple roots occurs in pir. Hence given cubic eqution hs two non-rel nd one rel roots. Illustrtion 38 : If q, r, s re positive, show tht the eqution ƒ () + q + r s = 0 hs one positive, one negtive nd two imginr roots. Product = s < 0 let roots be < 0 this is possible when - (i) (ii) (iii) one root is negtive & three re positive three roots re negtive & one is positive one root negtive, one positive & two roots imginr. ƒ () + q + r s As there is onl one chnge of sign, the eqution hs tmost one positive root. ƒ ( ) + q r s Agin there is onl one chnge of sign, the eqution hs tmost onl one negtive root. so (i), (ii) cn't be possible. Hence there is onl one negtive root, one positive root & two imginr roots.

J-Mthemtics Do ourself - : ( i ) Let be two of the roots of the eqution 3 p + q r = 0. If = 0, then show tht pq = r ( i i ) If two roots of 3 + 3 9 + c = 0 re equl, then find the vlue of c. (iii) If be the roots of 3 + b + c + d = 0, then find the vlue of () (b) 9. TR ANSFORM ATION OF TH QUATION : (c) ( ) Let + b + c = 0 be qudrtic eqution with two roots nd. If we hve to find n eqution whose roots re f() nd f(), i.e. some epression in & then this eqution cn be found b finding in terms of. Now s stisfies given eqution, put this in terms of directl in the eqution. = f () B trnsformtion, = g() (g()) + b(g()) + c = 0 This is the required eqution in. Illustrtion 39 : If the roots of + b + c = 0 re nd, then find the eqution whose roots re : () (),, put, = = b c 0 (b), Required eqution is c b + = 0 (b), put, = = (c), c b + = 0 Illustrtion 0 (c), Put = b c = 0 ( + c b) + ( c + b) +c = 0 Required eqution is ( + c b) + (b c) + c = 0 put = = + b + c = 0 b = + c + c + (c b ) + c = 0 Required eqution is + (c b ) + c = 0 :If the roots of 3 + b + c + d = 0 re,, then find eqution whose roots re = = d ( = d ) 50,,. NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY

J-Mthemtics Put = d Do ourself - 3 : d 3 + b d + c d + d = 0 Required eqution is d 3 bd + c = 0 ( i ) If re the roots of + b + c = 0, then find the eqution whose roots re (), (b), b b (c), ( i i ) If re roots of p + q = 0, then find the qudrtic eqution whose root re nd 3 3. Miscellneous Illustrtions : 3 3 ( )( ) Illustrtions : A polnomil in of degree greter thn three, leves reminders, nd when divided, respectivel, b ( ), ( + ) nd ( + ). Wht will be the reminder when it is divided b ( ) ( + ) ( + ). Let required polnomil be f() = p() ( ) ( + ) ( + ) + 0 + + B reminder theorem, f() =, f( ) =, f( ) =. 0 + + = 0 + = 0 + = Solving we get, 0 = 7 6, = 3, = 3 Reminder when f() is divided b ( ) ( + ) ( + ) will be 7 6 + 3 + 3. NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY Illustrtions :If re the roots of + p + q = 0, nd re the roots of + r + s = 0, evlute ( ) ( ) ( ) ( ) in terms of p, q, r nd s. Deduce the condition tht the equtions hve common root. re the roots of + p + q = 0 + = p, = q...() nd, re the roots of + r + s = 0 + = r, = s...() Now, ( ) ( ) ( ) ( ) = [ ( + ) + ] [ ( + ) + ] = ( + r + s) ( + r + s) = +r( + ) + r + s( + ) +sr( + ) + s = +r( + ) + r + s(( + ) )) + sr( + ) + s = q pqr + r q + s(p q) + sr ( p) + s = (q s) rpq + r q + sp prs = (q s) rq (p r) + sp (p r) = (q s) + (p r) (sp rq) For common root (Let = or = )...(3) then ( ) ( ) ( ) ( ) = 0...() from (3) nd (), we get (q s) + (p r) (sp rq) = 0 (q s) = (p r) (rq sp), which is the required condition. 5

J-Mthemtics Illustrtions 3 : If ( 5 + 3) ( + + ) < for ll R, then find the intervl in which lies. ( 5 + 3) ( + + ) <, R 5 + 3 < Let = P p + (p ) + p = 0 () Since is rel, (p ) p 0 p 3 () The minimum vlue of /( + + ) is. So, 5 + 3 < 5 + 5 < 0 5 5 5 5, ANSWRS FOR DO YOURSLF : ( i ) ( ), ; (b) ; (c) ± ; ( i i ), ; (iii) 7 3 ( i v ) 3, 5 : ( i ) b =, c = ; (ii) () imginr; (b) rel & distinct ; (c) rel & coincident 3 : ( i ) () c = 0; (b) c = ; (c) b negtive, c negtive : ( i ) 9 b, c 5 ; (ii) c = 0, 6 5 : ( i ) ( ) (, 3] [, ) ; (b) (, ) ; ( c ), 3 ; (d) ( 6, 3), {} (9, ) ; ( e ) [3,7]; ( f ) 6 : ( i ) (), = ; (b), = ( i i ) () (i) < 0 (ii) b < 0 (iii) c < 0 (iv) D > 0 (v) < 0 (vi) > 0 () (i) < 0 (ii) b > 0 (iii) c = 0 (iv) D > 0 (v) > 0 (vi) = 0 (3) (i) < 0 (ii) b = 0 (iii) c = 0 (iv) D = 0 (v) = 0 (vi) = 0 (iii) Third qudrnt ( i v ) () > 9/6 (b) < 7 : ( i ) (, ) (, 3/) ( i i ) R (0,] 8 : ( i ) () (,8) (b) (, /) ( i i ) (,) 9 : (ii) lest vlue = 0, gretest vlue =. 0 : ( i ) 3 < < 0 < < ; ( i i ) C ; (iii) < ; ( i v ) < ; (v) < < 6 : ( i ) 0, : ( i i ) 7, 5 ; (iii) () (b c), ( b ) 5 c, ( c ) (3d bc) d 3 : ( i ) ( ) c + (c b ) + = 0 ; ( b ) c b + = 0; ( c ) c + ( + c)b + ( + c) = 0 ( i i ) p(p 5p q + 5q ) + p q (p q)(p q) = 0 NOD6 ()\DATA\0\KOTA\J-ADVANCD\SMP\MATHS\UNIT#0\NG\0. QUADRATIC\.THORY