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Quadratic Equation– Session 3

Quadratic Equation– Session 3. Session Objective. 1. Condition for common root. 2. Set of solution of quadratic inequation. 3. Cubic equation. Condition for Common Root. _H007. The equations ax 2 + bx + c =  0 & a’x 2 + b’x + c’ =  0 has a common root(CR)  .

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Quadratic Equation– Session 3

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  1. Quadratic Equation– Session 3

  2. Session Objective 1. Condition for common root 2. Set of solution of quadratic inequation 3. Cubic equation

  3. Condition for Common Root _H007 The equations ax2 + bx + c =  0 & a’x2 + b’x + c’ =  0 has a common root(CR)  a 2 + b  + c =  0  a’ 2 + b’  + c’ =  0 Treating 2 and  as two different variable By rule of cross-multiplication ab’–ba’ bc’–cb’ ca’–ac’

  4. Condition for Common Root _H007 Condition for common root ofax2 + bx + c =  0 & a’x2 + b’x + c’ =  0 is (ca’-ac’)2=(bc’-cb’)(ab’-ba’)

  5. Illustrative Problem _H007 If x2-ax-21=0 and x2-3ax+35=0 (a>0)has a common root then value of a is (a)3 (b) 4 (c )2 (d) 4

  6. Illustrative Problem If x2-ax-21=0 and x2-3ax+35=0 (a>0)has a common root then value of a is (a)3 (b)4 (c )2 (d)4 _H007 Solution: Method 1 Let  be the common root 2-a -21=0 2-3a+35=0 By Cross- Multiplication

  7. Illustrative Problem _H007 If x2-ax-21=0 and x2-3ax+35=0 (a>0)has a common root then value of a is (a)3 (b)  4 (c )2 (d) 4 a2=16 a =  4 As a>0  a=4

  8. Illustrative Problem _H007 If x2-ax-21=0 and x2-3ax+35=0 (a>0)has a common root then value of a is (a)3 (b)  4 (c )2 (d) 4 Solution: Method 2 2- a - 21 = 0 ….(A) 2-3a+35 = 0 …..(B) (A) – (B)  2a = 56 Substituting ‘’ in (A) a =  4 As a>0  a=4

  9. Illustrative Problem _H007 If equation x2-ax+b=0 and x2+bx-a=0 has only one common root then prove that a-b=1 Solution: By observation at x=1 both the equation gives same value. x2- ax + b=0 …(A) L.H.S. = a-b-1 for x=1 x2 + bx - a=0 …(B) This means x=1 is the common root  a – b – 1= 0  a–b=1 Why? for x=1 both the equations give this

  10. Illustrative Problem _H007 If equation x2-ax+b=0 and x2+bx-a=0 has only one common root then prove that a-b=1 Solution: Method 2 • Let  be the common root then • 2 -a + b = 0 & • 2 + b - a = 0 subtracting one • from the other we get • (b + a)  - (b + a) = 0 •  = 1 provided b + a  0 Hence x = 1 is the common root  1 – a + b = 0 or a – b = 1 Why??

  11. Condition for Two Common Roots _H007 The equations ax2+bx+c=0  and a’x2+b’x+c’ = 0 have both roots common ax2 + bx + c  K(a’x2 + b’x + c’) why? when both the roots are common ,two equations will be same . But not necessarily identical. As x2–3x+2=0 and 2x2–6x+4=0 Same equation. Both have roots 1,2 But not identical For two roots to be common

  12. Illustrative Problem _H007 • If x2+ax+(a-2) = 0 and bx2+2x+ 6 = 0 have both roots common then a : b is • (a) 2 (b)1/2 (c) 4 (d)1/4 Solution: As both roots are common  a=-1

  13. ax2+bx+c > 0 ax2+bx+c  0 Quadratic Inequation _H009 A statement of inequality exist between L.H.S and R.H.S Quadratic Inequation If ax2+bx+c =0 has roots ,; let < ax2+bx+c = a(x- )(x- ) When ax2+bx+c >0 Let a>0  (x- )(x- )>0

  14. -  Number line   Quadratic Inequation _H009 (x- )(x- )>0 Either (x- )>0; (x- )>0 Or (x- )<0; (x- )<0  x >  and x>   x<<  x <  and x<   x>> x< x> • and  are not included in set of solutions for a(x- )(x- )>0 ; (a>0) x lies outside ,

  15. -  1/3 1/2 Illustrative Problem _H009 Find x for which 6x2-5x+1>0 holds true Solution: for 6 (x-1/3) (x-1/2)>0  Either x>1/2 or x<1/3

  16. -   Quadratic Inequation _H009 ax2+bx+c = a(x- )(x- ) and a>0 where  <  When ax2+bx+c < 0  (x- )(x- ) < 0 Either (x- )<0; (x- )>0 Or (x- )>0; (x- )<0  x >  and x<   x <  and x>   <x< No solution • and  are not included in set of solutions <x< for a(x- )(x- ) <0 ; (a>0) x lies within ,

  17. x 1/3 1/2 Illustrative Problem _H009 Find x for which 6x2-5x+1<0 holds true Solution: for 6 (x-1/3) (x-1/2)<0  1/3 <x< 1/2

  18. - -2 3 Illustrative Problem _H009 Solve for x : x2 - x – 6 > 0 Solution: Step1:factorize into linear terms (x-3) (x+2) >0 Step2 :Plot x for which x2-x–6=0 on number line x2-x–6 >0 for either x<-2 or x>3 As sign of a >0

  19. x x     Quadratic Inequation _H009 For a(x- )(x- )  0 x   x   Here set of solution contains , and all values outside , For a(x- )(x- )  0   x   x lies within , and also includes , in solution set

  20. Illustrative Problem _H009 Solution : x2 – 8x + 10  0 step1: Find the roots of the corresponding equation Roots of x2 – 8x + 10 = 0 are

  21. -  4-6 4+6 Illustrative Problem _H009 x2 – 8x + 10  0 Step2: Plot on number-line 4-6 and 4+6 are included in the solution set

  22. -  7 8 Illustrative Problem _H009 Solve for x : - x2 +15 x – 6 > 0 Solution: Here a=-1 Step1: Multiply the inequation with (-1)to make ‘a’ positive. Note- Corresponding sign of inequality will also change x2 –15 x + 6 < 0  (x-7)(x-8) <0 step2: Plot on Number line 7 <x<8

  23. Cubic Equation _H015 P(x)=ax3+bx2+cx+d A polynomial of degree 3 P(x)=0  ax3+bx2+cx+d=0 is a cubic equation when a  0 Number of roots of a cubic equation?

  24. Cubic Equation _H015 Let the roots of ax3+bx2+cx+d =0 be  ,,  ax3+bx2+cx+d a(x- ) (x-  ) (x- ) As ax2+bx+c has roots  , can be written as ax2+bx+c a(x- ) (x-  ) a[x3-(++)x2+(++)x-()]

  25. Cubic Equation _H015 ax3+bx2+cx+d a[x3-(++)x2+ (++)x-()] Comparing co-efficient

  26. Cubic Equation _H015 ax3+bx2+cx+d=0 a,b,c,dR 3 Maximum real root = ? As degree of equation is 3 Minimum real root? 0? Complex root occur in conjugate pair when co-efficient are real Maximum no of complex roots=2 Minimum no. of real root is 1

  27. Illustrative Problem _H015 If the roots of the equation x3-2x+4=0 are ,, then the value of (1+ ) (1+ ) (1+ ) is • 5 (b) –5 • (c )4 (d) None of these

  28. Illustrative Problem _H015 If the roots of the equation x3-2x+4=0 are ,, then the value of (1+ ) (1+ ) (1+ ) is (a)5 (b)–5 (c )4 (d) None of these Solution Method 1: x3-2x+4=0 has roots ,,  a=1, b=0, c=-2, d=4 (1+ ) (1+ ) (1+ )= 1+ + + =1-0-2-4 = -5

  29. Illustrative Problem _H015 If the roots of the equation x3-2x+4=0 are ,, then the value of (1+ ) (1+ ) (1+ ) is • 5 (b) –5 (c )4 (d) None of these Method 2 Let f(x)= x3-2x+4 = (x- ) (x- )(x- ) for x=-1 f(-1)= 5 = (-1- ) (-1- )(-1- ) (-1)3(1+ ) (1+ )(1+ ) = 5  (1+ ) (1+ )(1+ ) = -5

  30. Class -Exercise

  31. Class Exercise1 • If the equations ax2 + bx + c = 0 and cx2 + bx + a = 0 have one root common then • a + b + c = 0 • a + b – c = 0 • a – b + c = 0 • both (a) or (c) Solution: • By observation roots of one equation is • reciprocal to other. So both equation will have common root if it becomes 1 or –1

  32. Class Exercise1 • If the equations ax2 + bx + c = 0 and cx2 + bx + a = 0 have one root common then • a + b + c = 0 • a + b – c = 0 • a – b + c = 0 • both (a) and (c) When 1 is common root ,a + b + c = 0. when –1 is common root, a – b + c = 0

  33. Class Exercise2 • If x2 + ax + 3 = 0 and bx2 + 2x + 6 = 0 have both roots common then a : b is • (a) 2 (b)1/2 (c) 4 (d)1/4 Solution: As both roots are common

  34. Class Exercise3 Solution : x2 – 10x + 22  0 Factorize into linear terms by using perfect square method

  35. x 5-3 5+3 Class Exercise3 Plot on number-line 5-3 and 5-3 are included in the solution set

  36. Class Exercise4 • The number of integral values of x for which (x – 6) (x + 1) < 2 (x – 9) holds true are • (a) Two (b)Three (c) One (d) Zero Solution: (x – 6) (x + 1) < 2 (x – 9)  x2 – 7x + 12 < 0 or, x2 – 5x – 6 < 2x – 18 or, (x – 3) (x – 4) < 0  3 < x < 4 • So no integral values of x satisfies it.

  37. If , ,  are the roots of 2x3 + 3x2 – 2x + 1 = 0. Then the value of is • 17/4 (b) 41/4 • (c )9/4 (d) None of these Class Exercise5 2x3 + 3x2 – 2x + 1 = 0 Solution:

  38. Class Exercise5 If , ,  are the roots of 2x3 + 3x2 – 2x + 1 = 0. Then the value of is • 17/4 (b) 41/4 • (c )9/4 (d) None of these

  39. Class Exercise6 If ax2 + bx + c =  0 & bx2 + c x + a =  0 has a common root and a ‡ 0 then prove that a3+b3+c3=3abc Solution:  (bc – a2)2 = (ab – c2) (ac – b2)

  40. Class Exercise6 If ax2 + bx + c =  0 & bx2 + c x + a =  0 has a common root and a ‡ 0 then prove that a3+b3+c3=3abc  (bc – a2)2 = (ab – c2) (ac – b2) By expansion:  b2c2 + a4 – 2a2bc = a2bc – ab3 – ac3 + b2c2  a(a3 + b3 + c3) = 3a2bc a(a3 + b3 + c3 – 3abc) = 0 either a = 0 or a3 + b3 + c3 = 3abc As a  0 a3 + b3 + c3 = 3abc

  41. Class Exercise7 Find the cubic equation with real co-efficient whose two roots are given as 1 and (1 + i) Solution: • Imaginary roots occur in conjugate pair; when co-efficients are real Roots are 1, (1 – i) (1 + i) • Equation is or, (x – 1) (x2 – 2x + 2) = 0 x3 – 3x2 + 4x – 2 = 0

  42. Class Exercise8 If ax3 + bx2 + cx + d = 0 has roots , and  and , then find the equation whose roots are Solution: As roots are incremented by 2. • So desired equation can be found replacing x by (x– 2) • a(x – 2)3 + b(x – 2)2 + c(x – 2) + d = 0

  43. Class Exercise9 Find values of x for which the inequation (2x – 1) (x – 2) > (x – 3) (x – 4) holds true Solution: (2x – 1) (x – 2) >(x – 3) (x – 4)  2x2 – 5x + 2 > x2 – 7x + 12  x2 + 2x – 10 > 0

  44. Class Exercise9 Find values of x for which the inequation (2x – 1) (x – 2) > (x – 3) (x – 4) holds true

  45. Class Exercise10 • For what values of ‘a’ , • a(x-1)(x-2)>0 when 1 < x < 2 • (a) a > 0 (b) a < 0 (c) a = 0 (d) a = 1 Solution: When 1 < x < 2 (x – 1) > 0, (x – 2) < 0 As a (x – 1) (x – 2) > 0 a < 0

  46. Thank you

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