1 / 53

Quadratic Equation- Session1

Quadratic Equation- Session1. Session Objective. Definition of important terms (equation,expression,polynomial, identity,quadratic etc.). 2. Finding roots by factorization method. 3. General solution of roots. 4. Nature of roots.

gil-best
Télécharger la présentation

Quadratic Equation- Session1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Quadratic Equation- Session1

  2. Session Objective • Definition of important terms • (equation,expression,polynomial, identity,quadratic etc.) 2. Finding roots by factorization method 3. General solution of roots . 4. Nature of roots

  3. Quadratic Equation - Definitions (Expression & Equation) Expression: Representation of relationship between two (or more) variables _H001 Y= ax2+bx+c, Equation :Statement of equality between two expression ax2 + bx + c = 0  Root:-value(s) for which a equation satisfies Example: x2-4x+3 = 0  (x-3)(x-1) = 0 satisfies x2-4x+3 = 0 Roots of x2-4x+3 = 0  x = 3 or 1

  4. Quadratic EquationDefinitions (Polynomial) _H001 Polynomial : P(x) = a0 + a1x + a2x2 + … + anxn, where a0, a1, a2, … an are coefficients , and n is positive integer Degree of the polynomial : highest power of the variable A polynomial equation of degree n always have n roots Real or non-real

  5. Quadratic EquationDefinitions (Polynomial) _H001 Equation  2 roots (say 1,2) (x-1)(x-2)=0  x2 - 3x+2 = 0 2nd degree equation 2nd degree equation  2 roots • Roots are 1,2 (x- 1 )(x- 2 )=0 x2-(1+2)x+ 12= 0  ax2 + bx + c=0

  6. Quadratic EquationDefinitions (Polynomial) _H001 • Roots are 1,2,3 (x- 1 )(x- 2 ) (x- 3) =0  ax3 +bx2+cx+d = 0 3rd degree equation 3rd degree equation  3 roots • Roots are 1,2, 3,…….n (x- 1 )(x- 2 ) (x- 3)….. (x- n) =0  anxn+an-1xn-1+…….+ a0 =0 nth degree equation nth degree equation  n roots

  7. Quadratic EquationDefinitions (Quadratic & Roots) _H001 Quadratic: A polynomial of degree=2 y= ax2+bx+c ax2+bx+c = 0 is a quadratic equation. (a  0 ) A quadratic equation always has two roots

  8. Roots _H001 What are the roots of the equation (x+a)2=0 Where is the 2nd root of quadratic equation? Then what is its difference from x+a=0 x=-a ? (x+a)2=0 (x+a)(x+a) =0  x= -a, -a Also satisfies condition for quadratic equation two roots

  9. Identity _H001 Identity : Equation true for all values of the variable (x+1)2 = x2+2x+1 Equation holds true for all real x

  10. Polynomial identity _H001 If a polynomial equation of degree n satisfies for the values more than n it is an identity Example: (x-1)2 = x2-2x+1 Is a 2nd degree polynomial Satisfies for x=0 (0-1)2=0-0+1 Satisfies for x=1 (1-1)2=1-2+1 (-1-1)2=1+2+1 Satisfies for x=-1 2nd degree polynomial cannot have more than 2 roots (x-1)2 = x2-2x+1 is an identity

  11. Polynomial identity LO-H01 Polynomial of x If P(x)=Q(x) is an identity Co-efficient of like terms is same on both the side Illustrative example If (x+1)2=(a2)x2+2ax+a is an identity then find a?

  12. Illustrative Problem If (x+1)2=(a2)x2+2ax+a is an identity then find a? _H001 Solution (x+1)2=(a2)x2+2ax+a x2+2x+1 =(a2)x2+2ax+a is an identity Equating co-efficient  a= 1 x2 : a2=1 a=1 x : 2a=2 satisfies all equation constant: a=1

  13. Illustrative problem _H001 Find the roots of the following equation By observation Solution: For x=-a L.H.S= 0+0+1=1 = R.H.S = R.H.S For x=-b L.H.S= 0+1+0=1 For x=-c L.H.S= 1+0+0=1 = R.H.S

  14. Illustrative problem Find the roots of the following equation 2nd degree polynomial is satisfying for more than 2 values Its an identity Satisfies for all values of x i.e. on simplification the given equation becomes 0x2+0x+0=0

  15. -4,3 -2,6 4,-3 factors with opposite sign Quadratic Equation-Factorization Method _H002 Solve for x2+x-12=0 Step2: Sum of factors factors -1 product Step1: 4 -12 1 Step3: x2+(4-3)x -12=0  x2+4x-3x-12=0 (x+4)(x-3)=0 Roots are -4, 3

  16. Quadratic Equation-Factorization Method _H002  x2+(4-3)x -12=0 x2+x-12=0 (where roots are –4,3) Similarly if ax2+bx+c=0 has roots , ax2+bx+c  a(x2-(+)x + ) Comparing co-efficient of like terms:

  17. Properties of Roots _H005 Quadratic equation ax2+bx+c=0 , a,b,c R  and  The equation becomes: a { x2+ (b/a)x + (c/a) }= 0 ax2-(+ )x+  =0 a(x-)(x-)=0  x2-(sum) x+(product) =0

  18. Illustrative Problem _H002 Solve:- Solution: Step1:-Product a2-b2 Sum 1+a2-b2 Step2:-Factors 1, a2-b2 and (a+b), (a-b) 2a Step3:

  19. Illustrative Problem _H002 Solve: x2-2ax+a2-b2 = 0 Either {x-(a+b)}=0 or {x-(a-b)}=0 Ans : x=(a+b) ,(a-b)

  20. Illustrative Problem In a quadratic equation with leading co-efficient 1 , a student reads the co-efficient of x wrongly as 19 , which was actually 16 and obtains the roots as -15 and –4 . The correct roots are _H002 Hint:-Find constant term

  21. Illustrative Problem In a quadratic equation with leading co-efficient 1 , a student reads the co-efficient of x wrongly as 19 , which was actually 16 and obtains the roots as -15 and –4 . The correct roots are _H002 Solution: Step 1: equation of roots –15 & -4 (x+15)(x+4)=0 Or x2 +19x+60=0 Step2: Get the original equation x2+16x+60=0 Roots are –10 & -6

  22. Illustrative Problem _H005 Product of the roots of the equation x2+6x+ 2+1=0 is –2,Then the value of  is (a)-2, (b)-1, (c)2, (d)1 [DCE-1999]

  23. Illustrative Problem Product of the roots of the equation x2+6x+ 2+1=0 is –2,Then the value of  is (a)-2, (b)-1, (c)2, (d)1 _H005 x2+6x+ 2+1=0 Product of the roots  (2+1)/=-2 (+1)2=0  =-1

  24. General Solution _H003 To find roots of ax2 + bx + c = 0 Step 1: Convert it in perfect square term • Multiplying this equation by 4a, • 4a2x2 + 4abx + 4ac = 0 HOW !! • Add and subtract b2 • (4a2x2 + 4abx + b2) + 4ac - b2 = 0 (2ax + b)2 = b2 - 4ac

  25. General Solution _H003 Step 2: Solve For x ax2 + bx + c = 0 has two roots as

  26. General Solution _H003 (b2 - 4ac)discriminant of the quadratic equation, and is denoted by D . Roots are This is called the general solution of a quadratic equation

  27. Illustrative Problem _H003 Find the roots of the equation x2-2x-1=0 by factorization method Solution: As middle term cannot be splitted form the square involving terms of x x2-2x-1=0 (x2-2x+1) –2=0 (x-1)2-(2)2=0 Form linear factors (x-1+ 2) (x-1- 2)=0 Roots are : 1+2, 1-2

  28. Ans: Roots are Illustrative Problem _H003 Find the roots of the equation x2-10x+22=0 Solution: Here a=1, b=-10, c=22 Apply the general solution form

  29. D > 0 is real  (D is not a perfect square) (D is perfect square) D < 0 is not real  Nature of Roots _H004 Discriminant, D=b2-4ac Roots are real a, b, c are rational Rational Irrational D = 0 Roots are real and equal Roots are imaginary

  30. Illustrative Problem _H004 Find the nature of the roots of the equation x2+2(3a+5)x+2(9a2+25)=0 Solution: D=4(3a+5)2-4.2(9a2+5) = -36a2+120a-100 =-4(3a-5)2 D<0 As (3a-5)2 >0 except a=5/3 Roots are imaginary except a=5/3

  31. Irrational Roots Occur in Pair _H004 ax2 + bx + c =  0 ,a,b,c Rational Irrational when Q is not perfect square rational  = P+ Q and = P- Q Irrational roots occur in conjugate pair when co-efficient are rational

  32. Complex Roots Occur in Pair _H004 In ax2 + bx + c =  0 ,a,b,c Real If one root complex (p+iq) Other its complex conjugate (p-iq ) Prove yourself In quadratic equation with real co-eff complex roots occur in conjugate pair

  33. Illustrative Problem _H004 Find the quadratic equation with rational co-eff having a root 3+5 Solution: One root (3+5)  other root (3-5) Required equation (x-{3+5})(x- {3-5})=0 x2-{(3+5)+(3-5)}x+(3+5) (3-5) =0 Ans: x2-6x+4=0

  34. Illustrative Problem _H004 • If the roots of the equation • (b-x)2 -4(a-x)(c-x)=0 • are equal then • b2=ac (b)a=b=c • (c)a=2b=c (d) None of these

  35. Illustrative Problem _H004 • If the roots of the equation • (b-x)2 -4(a-x)(c-x)=0 are equal then • b2=ac (b)a=b=c • (c)a=2b=c (d) None of these Solution: (b-x)2 -4(a-x)(c-x)=0 x2+b2-2bx-4{x2-(a+c)x+ac}=0 3x2+2x(b-2a-2c)+(4ac-b2)=0 Roots are equal  D=0 D=4(b-2a-2c)2-4.3.(4ac-b2)=0  b2+4a2+4c2-4ab-4bc+8ac-12ac+3b2=0 4(a2+b2+c2-ab-bc-ca)=0

  36. Illustrative Problem _H004 • If the roots of the equation • (b-x)2 -4(a-x)(c-x)=0 are equal then • b2=ac (b)a=b=c • (c)a=2b=c (d) None of these 4(a2+b2+c2-ab-bc-ca)=0 Sum of 3 square is zero How/When it’s possible? (a-b)2+(b-c)2+(c-a)2=0 It’s only possible when each separately be zero a-b=0; b-c=0 ; and c-a=0 a=b=c

  37. Illustrative Problem _H004 • For what values of k • (4-k)x2+(2k+4)x+(8k+1) becomes • a perfect square • 3 or 0 (b) 4 or 0 • (c ) 3 or 4 (d) None of these Hint: (4-k)x2+(2k+4)x+(8k+1) becomes a perfect square Roots of the corresponding equation are equal

  38. Illustrative Problem _H004 • For what values of k • (4-k)x2+(2k+4)x+(8k+1) becomes • a perfect square • 3 or 0 (b) 4 or 0 • (c ) 3 or 4 (d) None of these  (4-k)x2+(2k+4)x+(8k+1) =0 has equal roots D = (2k+4)2-4.(4-k).(8k+1)=0  4k2+16k+16-4(31k-8k2+4)=0  k2+4k+4+8k2-31k-4=0 9k2-27k=0 k=0 or 3

  39. Class Exercise1 Number of roots of the equation (x + 1)3 – (x – 1)3 = 0 are (a) two (b) three (c) four (d) None of these (x + 1)3 –(x –1)3 = 0 Solution: 6x2 +2 = 0 2(3x2 +1) = 0, It is a quadratic equation  must have two roots.

  40. Class Exercise2 (x + 1)3 = K2x3 +(K+2) x2 + (a – 2 )x + b is an identity, Then value of (K, a, b) are (a) (-1, 5, 1) (b) (1,5,1) (c) (1, 5 ,1) (d) None of these Solution: Since (x + 1)3 = K2x3 + (K+2) x2 +(a –2)x + b is an identity, co-efficient of like terms of both the sides are the same x3 + 3x2 + 3x + 1 =K2x3 + (K+2) x2 +(a –2)x + b K2=1-------(i) K+2=3---(ii)

  41. Class Exercise2 (x + 1)3 = K2x3 +(K+2) x2 + (a – 2 )x + b is an identity, Then value of (K, a, b) are (a) (-1, 5, 1) (b) (1,5,1) (c) (1, 5 ,1) (d) None of these K2=1-------(i) K+2=3---(ii) K=1 a–2 = 3  a=5 b = 1

  42. Class Exercise3 • Roots of the equation cx2 – cx + c + bx2 – cx – b = 0 are • c and b (b)1 , • (c) (c + b) and (c – b) (d) None of these Solution: (c + b)x2 – 2cx + (c – b) = 0  (c+b)x2–{(c+b)+(c–b)}x+(c–b)=0 Roots are 1 and  (c+b)x2–(c+b)x–(c–b)x+(c –b)= 0  (c+b)x (x – 1) – (c – b) (x – 1) = 0  (x – 1) {(c+b)x –(c – b)} = 0

  43. Class Exercise4 Let , are the roots of the equation (x-a)(x-b)=c, c 0. Then roots of the equation (x- )(x- )+c = 0 are (a) a,c (b)b,c (c ) a,b (d)(a+c),(b+c) (x-a)(x-b)=c , are the roots  x2-(a+b)x+ ab-c=0 So +=(a+b);  =ab-c……(1) Now (x-)(x- )+c = 0 x2-(+ )x+ +c=0 x2-(a+b )x+ ab=0 by(1) Roots are a and b (x-a) (x-b)=0

  44. Class Exercise5 • 5.The equation which has 5+3 and 4+2 as the only roots is • never possible • (b) a quadratic equation with rational co-efficient • (c) a quadratic equation with irrational co-efficient • (d) not a quadratic equation Solution: Since it has two roots it is a quadratic equation. As irrational roots are not in conjugate form. Co-efficient are not rational.

  45. If the sum of the roots of is zero, then prove that product of the roots is . Class Exercise6 Solution: c[(x + a) + (x + b)] = (x + a) (x + b) 2cx + (a + b) c = x2 + (a + b) x + ab x2 + (a + b – 2c) x + (ab – ac – bc) = 0 As sum of roots = 0  a + b = 2c Product of roots = ab – ac – bc

  46. = ab- Class Exercise6 If the sum of the roots of is zero, then prove that product of the roots is . Sum of roots = 0  a + b = 2c Product of roots = ab – ac – bc = ab – c (a + b)

  47. Class Exercise7 Both the roots of the equation (x–b)(x–c)+(x–c)(x–a)+(x–b)(x–a)=0 are always :a,b,c,R (a) Equal (b) Imaginary (c) Real (d) Rational Solution: (x – b) (x – c) + (x – c) (x – a) + (x – b) (x – a) = 0 or 3x2 – 2(a + b + c) x + (ab + bc + ca) = 0 D = 4 (a + b + c)2 – 4.3.(ab + bc + ca) = 4 [(a + b + c)2 – 3(ab + bc + ca)]

  48. Class Exercise7 Both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – b) (x – a) = 0 are always :a,b,c,R (a) Equal (b) Imaginary (c) Real (d) Rational D= 4 [(a + b + c)2 – 3(ab + bc + ca)] = 4 (a2 + b2 + c2 – bc – ca – ab) =2[(a-b)2+(b-c)2+(c-a)2] As sum of square quantities are always positive; D > 0 Roots are real.

  49. Class Exercise8 The roots of the equation (a+b+c)x2–2(a+b)x+(a+b–c)=0 are (given that a, b, c are rational.) (a) Real and equal (b) Rational (c) Imaginary (d) None of these Solution: Sum of the co-efficient is zero.  (a + b + c) 12 + 2 (a + b).1 + (a + b – c) = 0  1 is a root, which is rational  so other root will be rational.

  50. If the roots of the equation (a2+b2)x2–2(ac+bd)x+(c2+d2)=0 are equal then prove that Class Exercise 9 Solution: D = 0 4 (ac+bd)2- 4 (a2+b2)(c2+d2)= 0 a2c2 + b2d2 + 2abcd = (a2 + b2) (c2 + d2)  2abcd = a2d2 + b2c2  a2d2 + b2c2 – 2abcd = 0  (ad – bc)2 = 0  ad – bc = 0  ad = bc

More Related