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

Quadratic Equation - Definitions (Expression & Equation) 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 Roots of x2-4x+3 = 0 satisfies x2-4x+3 = 0  x = 3 or 1

Quadratic Equation Definitions (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

Quadratic Equation Definitions (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

Quadratic Equation Definitions (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

Quadratic Equation Definitions (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

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

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

_H001 Polynomial identity 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 Satisfies for x=-1 (-1-1)2=1+2+1 2nd degree polynomial cannot have more than 2 roots (x-1)2 = x2-2x+1 is an identity

LO-H01 Polynomial identity 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?

_H001 Illustrative Problem Solution (x+1)2=(a2)x2+2ax+a 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 x2 : a2=1  a= 1 a=1 x : 2a=2 satisfies all equation constant: a=1

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

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

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

Quadratic Equation -Factorization Method x2+x-12=0  x2+(4-3)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:

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

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

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

_H002 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

_H002 Illustrative Problem Solution: 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

_H005 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 [DCE-1999]

_H005 Illustrative Problem  =-1 x2+6x+ 2+1=0 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

_H003 General Solution 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

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

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

_H003 Illustrative Problem 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

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

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

Roots are imaginary except a=5/3 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

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

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

_H004 Illustrative Problem 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

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

_H004 Illustrative Problem Solution: (b-x)2 -4(a-x)(c-x)=0 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

_H004 Illustrative Problem 4(a2+b2+c2-ab-bc-ca)=0 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 a-b=0; b-c=0 ; and c-a=0 It’s only possible when each separately be zero a=b=c

_H004 Illustrative Problem 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

_H004 Illustrative Problem  (4-k)x2+(2k+4)x+(8k+1) =0 has equal roots 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

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 Solution: (x + 1)3 – (x –1)3 = 0 6x2 +2 = 0 2(3x2 +1) = 0, It is a quadratic equation  must have two roots.

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)

Class Exercise2 K2=1-------(i) K+2=3---(ii) K=1 a–2 = 3  a=5 b = 1 (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

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

Class Exercise4 (x-a)(x-b)=c , are the roots  x2-(a+b)x+ ab-c=0 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 (x-a)(x-b)-c=(x-a)(x-b) (x-a)(x-b)=(x-a)(x-b)+c x2-(+ )x+ +c=0 x2-(a+b )x+ ab=0 by(1) (x-a) (x-b)=0 Roots are a and b

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.

Class Exercise6 If the sum of the roots of is zero, then prove that product of the roots is . 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

Class Exercise6 Sum of roots = 0  a + b = 2c 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) = ab-

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)]

Class Exercise7 D= 4 [(a + b + c)2 – 3(ab + bc + ca)] 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.

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.

Class Exercise 9 If the roots of the equation (a2+b2)x2–2(ac+bd)x+(c2+d2)=0 are equal then prove that 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

Class Exercise10 If the equation ax + by = 1 and cx2 + dy2 = 1 have only one solution, then prove that and Solution: what is geomatrical significance of this ax + by = 1  y = (1 – ax) ... (i) cx2 + dy2 = 1 or cx2 + d (1 – ax)2 = 1

Class Exercise10 or, b2 cx2 + d (a2x2 – 2ax + 1) = b2 If the equation ax + by = 1 and cx2 + dy2 = 1 have only one solution, then prove that and or, b2 cx2 + d (a2x2 – 2ax + 1) = b2 or x2 (b2c + a2d) – 2adx + (d – b2) = 0 ... (ii) As there is only one root D = 0  4a2d2 – 4(b2c + a2d) (d – b2) = 0 or a2d2 – (b2dc – b4c + a2d2 – a2b2d) = 0 or b4c – b2dc + a2b2d = 0

Class Exercise10 b4c – b2dc + a2b2d = 0 or b4c – b2dc + a2b2d = 0 If the equation ax + by = 1 and cx2 + dy2 = 1 have only one solution, then prove that and b4c – b2dc + a2b2d = 0 or b4c – b2dc + a2b2d = 0 [Dividing both sides by b2dc] when D=0;value of x from (ii) By using (i) and (iii), y=b/d