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1 THEORY OF EQUATIONS OBJECTIVE PROBLEMS. If the eqution x 6x 0 0 ) - ) 4) -. If the sum of two oots of the eqution k is -48 ) 6 ) 48 4) 4. If the poduct of two oots of 4 ) -4 ) 4) - 4. If one oot of is + + = hs one imginy oot +i, then its el oot is x 5x kx 4 0 x x 6x k 0 x 5x kx = is zeo, then vlue of + = is, then k = + = is the ecipocl of nothe, then the vlue of k 5 ) 4 ) 4) 5. If two oots of the condition is ) p q = e connected by the eltion αβ + = 0, then x px qx = ) p p q = 4) 6. If two of the oots of 7 4q 0 + = ) 7. If one of the oots of two, then k = x qx 0 + p + q + = 0 q + p + q + = = e equl, the condition is 4 7q 0 = ) x 6x 5x k 0 - ) ) 4) - 8. If α, β, γ e the oots of β + γ, γ + α, α + β is 7 4q 0 + = 4) 4 + 7q = = is equl to hlf the sum of the othe x px qx 0 + =, then the eqution whose oots e x px + ( p + q) x + ( qp) = 0 ) x + px + p q x + + qp = 0 ) x + px + ( p + q) x + ( + qp) = 0 4) x + px p + q x + qp = 0
2 9. If the poduct of the two oots of 4 x + px + qx + x + s = 0 is equl to the poduct of the othe two, then ) ps ps = ) = 4) p s = p s = 0. The condition tht the eqution mgnitude but opposite in sign is x px qx 0 + = my hve two oots equl in pq+=0 ) pq-=0 ) pq-=0 4) pq-=0. If the sum of two oots of ) ) 4). If α, β, γ e the oots of ) p q + q p ) 4). If α, β, γ e the oots of 8 ) 4 ) 4. If α, β, γ e the oots of ) q q + p p p q x px qx 0 + =, then pq = + =, then + + = α β β γ γ α x px qx 0 q + p + + =, then + + = β γ γ α α β x x x ) 4 + =, then + + = α β γ x px qx 0 ) 4) p q q + p
3 5. If the oots of,, 6 ), 6, ) 6. The oots of x 9x + 0x = 0 e in the tio :, then the oots e,, x 9x + x = 0 e 4),, 5 ± i ), ± i ), ± i 4), ± i 7. If,, α, β e the oots of α =, β = ) α =, β = ) α =, β = 4) α =, β = x x 5x The eqution whose oots e -, +i is ) x x x = ) x x x = 4) + + =, then x x + x 5 = 0 x + x + x + 5 = 0 9. The cubic eqution hving, the oots +, is ) ) 4) x + 5x + 5x + = 0 x 5x + 5x = 0 x 5x 5x = 0 x 5x 5x + = 0 0. The oots of the eqution ) p pq ) + = 4) p pq. If the oots of the eqution is ) b bc d 0 x px qx = e in A.P. then p pq = = p pq + = ) b + bc d = 0 4) x bx cx d = e in A.P., then the condition b + bc + d = 0 b bc d = 0. If the oots of the eqution, 4, 8 ),/,/4 ),,4 4),5,4 x 7x + 4x 8 = 0 e in G.P., then the oots e
4 . If the oots of the eqution,, 9 ) 4. If the oots of,, ),, 4 8x 4x + 7x = 0 e in G.P., then the oots e 4),,4 x + x + kx + 8 = 0 e in G.P., the vlue of k is ) - ) 4) - 5. The oots of the eqution x px qx 0 p = q ) p = q ) p 6. If the oots of the eqution men oot is q ) q 7. If the oots of ) q 7 9pq ) p q x px qx = e in G.P. Then x px qx 0 = q 4) p = q + = e in H.P., then the vlue of the 4) q p = e in H.P., then + = ) = 4) q 7 9q 8. The oots of the eqution b c c b ) c ) 4) b 9. If α, β, γ e the oots of eqution ) x 5x 5x = ) x 5x 5x 6 0 = 4) + = q 7 9pq = q 7 9pq x x bx c 0 + = e in H.P. then the men oot is + =, then x x 8x 6 0 x 5x + 5x 0 = = x 5x 5x 6 0 α, β, γ e the oots of the 0. If α, β, γ e the oots of x x + 5x = 0, then the eqution whose oots e βγ +, γα + αβ + α β γ is ) x + 0x + x + 64 = 0 ) x + 0x + x 64 = 0 4) x 0x + x 64 = 0 x 0x x + 64 = 0
5 . If α, β, γ e the oots of the eqution x + qx + = 0, the eqution whose oots β + γ γ + α α + β α β γ e,, is ) x + qx = 0 ) x qx 0 + = 4) x qx = 0 x + qx + = 0. If f ( x) 5x x x + 7 = 0, then ) 5x 7x 4x = ) 5x 7x 4x = 4). The eqution whose oots e the oots of diminished by 4 is ) 4 x x 4x 55x = ) 4 x x 4x 55x = 4) f x expessed in powe seies of (x-) is 5x + 7x 4x 9 = 0 5x 7x + 4x 9 = 0 4 x 5x 7x 7x = ech 4 x + x 4x + 55x + 9 = 0 4 x x 4x 55x + 9 = 0 4. The eqution whose oots e k times the oots of x x + x x + = 0 is 6 8 n eqution with integl coefficients. Then k = ) ) 5 4) 6 5. If -4,-,- e the oots of x 6x 9x 4 0 x + 6 x + 9 x + 4 = 0 e 4 5 5,, ) ) 4, 5, 5 4) =, then the oots of the eqution 0,, 4 5 5,, 6. The eqution whose oots e less by thn the oots of ) x x 0 + = ) x + x + = 0 4) 4x 8x 0 = 0 x x = 0 x 4x = is
6 7. If,, e the oots of 6x x + 9x = 0 e,, ),, x 9x + x 6 = 0, then the oots of the eqution ),, 4),, 8. The second tem of the eqution diminishing its oots by ) - ) 4) - x 6x x 0 9. By emoving the second tem in the eqution tnsfomed eqution is ) y 9y = ) y 9y = 4) 40. If y 9y + 6 = 0 y 9y 6 = = cn be emoved by x x x = the f x = x x + 4 = 0 hs epeted oot, then tht oot is ) - ) 4) - 4. The multiple oot of x x x 0 + = is ) - ) - 4) 4. If the eqution q = ) q x qx 0 + = hs two equl oots, then = ) 4. If α, β, γ e the oots of q = 4) q = =, then = α β x 4x 5x 0 4 ) ) 4 4) 44. If α, β, γ e the oots of x + bx + cx + d = 0, then the vlue of α β = c + bd ) bd c ) c bd 4) c + bd
7 45. If α, β, γ e the oots of x 9x 6x =, then the vlue of Σ( α + )( β + ) is 07 ) 08 ) 8 4) The oots of x x + 6x 6 = 0 e in A.P. ) G.P. ) A.G.P 4) H.P. 47. The numbe of el oots of eqution ) ) 6 4) If α is n imginy oot of nd ) α + α is x x 0 = ) x x 0 + = 4) 5 x = 0 x + x = 0 x + x + = 0 x + / x + x + / x = 0 is, then the eqution whose oots e α + α 4
8 . Ans () THEORY OF EQUATIONS HINTS AND SOLUTIONS Given +i is one oot -i will lso be oot. Sum of the two oots = +i+-i= s = sum of ll the oots = 0. The thid oot = 0- = -. Ans.() Let the oots be,, s α α β = α + β = = β = Since is oot of the given eqution,. Ans () 6 + k = 0 k = 48 Let α, β, γ be the oots so tht αβ =. s = αβγ = 4 γ = 4 γ = k = = 4 k = 4. Ans.( Let the oots be α,, β α s = α β = ( 4) = 4 β = 4 α 64-5(6)+4k-4=0 4k = 0 k = 5 5. Ans.(). But β is oot of the G.E. Let the oots be α, β, γ. Then αβγ =. Given αβ =. γ = γ =. But γ is oot of G.E. + p + q + = 0 + pq + q + = 0 is the condition. 6. Ans ( Let the oots be α, α, β. Then α + α + β = 0 β = α
9 α. α. β = α β = α = α = But α is oot of G.E. α qα + = 0 + q + = qα = q α = q = 8 8 Requied condition is 7. Ans () 4q + 7 = 0 4q + 7 = 0 Given one oot is equl to hlf the sum of the othe two. The oots e in A.P. Let the oots be c, + d f = 0 6 s = d d = = = = (- is oot of f(x) = k = 0 k = 8. Ans ( Given Σα = p, Σαβ = q αβγ = = β + γ = α + β + γ α = α = Let y p p x Requied eqution is α = x x = p y) p y p p y + q p y = 0 y py + p + q y + pq = 0 p y p y + py p py + p y + pq qy = = 9. Ans () x px p q x pq 0 Given Σα = p, Σαβ = q, Σαβγ =, αβγδ = s. αβ αβ = s α β = s. Also given αβ = γδ. Now αβγ + αβδ + αγδ + βγδ = αβ( γ + δ ) + γδ( α + β ) = αβ( α + β + γ + δ ) = ( γδ = αβ) αβ( p) = α β p = sp =.
10 0. Ans () Let the oots be α, α, β s = α α + β = p = p β = p x = p But β is oot of G.E. p p.p + qp = 0 pq = 0. Ans () Let the oots be α, αβ S = α α + β = p β = p But β is oot of the given eqution β pβ + qβ = 0 p p p + qp = 0 pq 0 pq. Ans () Given α +β+ γ = p, αβ +βγ + γα = q, αβγ = G.E.. Ans () γ + α + β = = α β γ ( α + β + γ) ( αβ + βγ + γα) ( αβγ) Given α + β + γ =, αβ + βγ + γα =. αβγ = 4 α + β + γ + + = β γ γ α α β α β γ = α + β + γ ( αβ + βγ + γα ) ( αβγ) = =. = 4 = = = p q β γ + γ α + α β 4. Ans () α + β + γ = p, αβ + βγ + γαq, αβγ = + + = α β γ α β γ ( βγ + γα + αβ) αβγ ( α + β + γ) ( αβγ) 5. Ans ( Let the oots be α, α, β. α + α +β = 0...( q = β = 9 4α, Fom () : α ( 9 4α ) = α. α. β =...() p
11 9α 4α = 4 4α 9α 4 = 0. By inspection : α = β = 9 8 = Hence, the oots e,6,,,6 6. Ans () By inspection x = is oot of the eqution. ( x ) is fcto of x x + x + = + = x x x x x 0 x =, x = ± i 7. Ans(4) + α +β =. α + β = αβ = 6 α nd ( α ) = 6 α α = α α + = α = 6 0 0,, β =, α =, β = 8. Ans () Given -, +i e the oots of given eqution. i is lso oot of the given eqution. Requied eqution is (x+ (x--i) (x-+i) = 0 x + x + = 0 x + x 4x + 5 = = 9. Ans () x x x 5 0 Itionl oots occu in pis is lso oot. The cubic eqution is ( x )( x + )( x = 0 x ) x = 0 x 4x + x = 0 x 5x + 5x = 0
12 0. Ans() Let the oots be d,, +d. d d = p = p = p But = p + p p + q p + = = p p pq 0 p + = pq is the equied condition.. Ans ( Given eqution is b c d =...( x x x 0 Let the oots be α β, α nd α + β. b b b sα β + α + α + β = α = α = Since α is oot of (, we hve b b b b d = b bc d + = 0 b bc + d = 0is the condition.. Ans() Let the oots be,, s = ( )( ) = ( 8) = 8 = 8 = s = + + = ( 7) = = = 0 = 0 o / The oots e,,4.. Ans() = Given =...(let Let the oots of ( be x x x 0 s =.. = = 8 = = 8,,
13 4 4 7 s = + + = = + + = = + = ( ( ) = 0 = The oots be 4. Ans (,, = = 8 = o. But is oot of the given eqution k + 8 = 0 k = 4 k = 5. Ans (4) Let the oots be,, = = = / s But is oot + p + q + = 0 / = / p q 0 q p = q p = q 6. Ans () / / Let the oots be,, d d ( d) ( d)( + d) ( + d) p = q p = q p = q / / /.. = = + d + d ( d) ( + d) + + = q + d + + d q = q = q = d d ( ) ( + ) Men oot is = q
14 7. Ans (,, Let d + d be the oots s = =, s = + + = q ( d) ( + d) ( d) ( d)( + d) ( + d) + d + + d = q ( d) ( + d) ( ) = q = q But is oot + q. + = p. + q. 0 + = q q q + + = 7 9pq q q = 7 9pq q q 0 + = + = 8. Ans (4) 7 9pq q 0 q 7 9pq Put x = /y in G.E. Then cy by y 0 + =...( The oots of () e in A.P. let α β, α, α +β be the oots of ( b b b b s = α β + α + β = α = α = α = c c c c b c The men oot of ( c. The men oot of G.E. is b. 9. Ans(4) Given α + β + γ =, αβ +βγ + γα = 8αβγ = 6 Requied eqution is x x x 0 α + β + γ + α β + γ α +β γ α β γ = = x x α + β + γ Σαβ + x Σαβ αβγσα αβγ = 0 + = x x 6 x x + 5x + 5x 6 = 0
15 0. Ans. () Given Σα =, Σαβ = 5, αβγ = Let y = αβγ βγ + = = = = α α α α x ( ) = x x = 4 y Requied eqution is + = 0 y y y + = 64 0y y 0 y 0y + y 64 = 0 o. Ans () Given Σα = 0, Σαβ = q, αβγ = β + γ α + β + γ 0 α α α α α α Let y = = = = = = x = α x y x 0x 64 0 =. Putting this vlue in G.E., we hve: + q 0 + = y y y qy = 0 o. Ans () Given x qx = 0 f x = 5x x x + 7 = 0 f x = 0 We hve to diminish the oots by Diminishing the oots of ( by (), we get The tnsfomed eqution is 5x + 7x 4x 9 = 0
16 . Ans () The tnsfomed eqution is 4. Ans (6) G.E is x x + x x + = x x + x x + = 0 0 Multiplying the oots of ( by 4 x + x + 4x + 55x 9 = 0...(. = 6, we emove the fctionl coefficients. k = 6 5. Ans () Diminishing the oots of the given eqution by -/ we get the second eqution. The oots of the second eqution e : + 4 +, +, + 0,, 6. Ans () Diminish the oots of G.E. by, we hve: Tnsfomed eqution is x + x + = 0.
17 7. Ans ( G.E. is x x x x 9x + x 6 = 0. Put /x in the plce of x = 0 oots e /,,. 6x x + 9x = 0 is the ecipocl eqution of G.E. whose Hence, the oots of tnsfomed eqution /,, /. 8. Ans (4) G.E. is + + = whee o =, = 6, =, = x 6x x 0 To emove the second tem, diminish the oots by h = 9. Ans ( Hee 0 = =, h = = = n0 Diminishing the oots by to emove the second tem The tnsfomed eqution is 40. Ans ( f x = x x + 4 Given y + 9y + 6 = 0 f x = x 6x = x x = 0 x = 0, 0 6 = = n f = = 0 f. But ( 0) 0 is epeted oot. 0
18 4. Ans ( Let f x = x x x + f x = x x f ( x) 0 x x 0 = = x x + = 0 x =, / But f( = 0 nd f ( / ) 0. is epeted oot. 4. Ans (4) Let the oots be α, α, β. Then α + α + β = 0 α. α. β = β = α, α ( α ) = α = 4. Ans ( α + β + γ = 4, αβ + βγ + γα = 5, αβγ = γ + α + β Σ = + + = α β α β β γ γ α α β γ = α + β + γ ( αβ + βγ + γα ) ( αβγ) 44. Ans () G.E. is x x x 0 ( ) b c d =...( b c d Σα = Σαβ =, αβγ = = = = 4 Σα β = α β + β γ + γ α = ( αβ + βγ + γα) αβγ ( α + β + γ) = 45. Ans ( = = c d b c bd Let y = α + = x + x = y. The tnsfomed eqution is y 9 y + 6 y 4 = 0 y + 7 9y + 7y 9y 8+ 54y + 6y 78 4 = 0
19 y 8y + 07y 0 = 0 Whose oots e α +, β +, γ + s = Σ α + β + = Ans () x x + 6x 6 [ ] = x 6 x x 6 = 0 x 6 x + 6x + 6 x = 0 x 6 x 5x + 6 = 0 x 6 x x = 0 x =, 6, These oots e in G.P. 47. Ans (4) Given eq. is x + + x + = 0 x x x + x + + = 0 x + + = 0 x + = 0( o) x + = 0 (O) x x x x x + + = 0 x Thee e no el oots. 48. Ans ()
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