1 / 53

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.

Télécharger la présentation

E N D

### Presentation Transcript

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