Quadratic Equation

Quadratic Equation POLYNOMIAL Algebraic expression containing many terms is called Polynomial. f(x) = 3x – 4 is single degree polynomial or Linear polynomial. e.g4x4 + 3x3 – 7x2 + 5x + 3, 3x3 + x2 – 3x + 5 f(x) = bx is odd Linear polynomial f(x) = a0 + a1x + a2x2 + a3x3 +..... an–1 x n–1 +anxn where x is a variable, a0,a1, a2 ..........anC. 1. 2. QUADRATIC EXPRESSION Real Polynomial : Let a0, a1,a2.....an be real numbers and x is a real variable. A Polynomial of degree two of the form ax2 + bx + c (a  0 ) is called a quadratic expression in x. Then f(x) = a0 + a1 x + a2 x2 +..... + an xn is called real polynomial of real variable x with real coefficients. e.g 3x2 + 7x + 5, x2 – 7x + 3 eg. – 3x3 – 4x2 + 5x – 4, x2– 2x + 1 etc. are real polynomials. General form : – f(x) where a, b, c  C and Complex Polynomial: If a0,a1,a2...an be complex numbers and x is a varying complex number, then f(x) = a0+ a1x + a2x2 +...... anxn is called a complex polynomial of complex variable x with complex coefficients. eg.- 3x 2 – (2+ 4 i) x + (5i–4), x 3 –5ix 2 + (1+2i) x+4 etc. are complex polynomials. 3. f(x) = 4x3 + 3x2 – 7x + 5 is 3 degree polynomial Degree of Polynomial : Highest Power of variable x in a polynomial is called as a degree of polynomial. e.g. f(x)=a0+a1x+a2x2+a3x3+....an–1xn–1+anxn is n degree polynomial. = ax2 + bx + c a0 QUADRATIC EQAUTION An equation ax2 + bx + c = 0 (where a  0, and a,b,c R), is called a quadratic equation. Here a, b and c are called coefficient of the equation. This equation always has two roots. Let the roots be  and . 1. Roots of a Quadratic Equation The values of variable x which satisfy the quadratic equation is called as Roots (also called solutions or zeros) of a Quadratic Equation. 27 SOLUTION OF QUADRATIC EQUATION 1. Ex.2 Factorization Method : Let ax2 + bx + c = a(x–) (x –) = 0 Then x =  and x =  will satisfy the given equation. Hence factorize the equation and equating each to zero gives roots of equation. e.g. 3x2 – 2x –1 x 2. Quadratic Equation The roots of the equation x2 – 4x + 1 = 0 are (A) 2 (B) 2, 4 3 (C) – 2 (D) 3 3 2 Here a = 1, b = 4, c = 1 Sol. Using Hindu Method = 0 (x – 1)( 3x + 1) = 0 x= 1 3 =1, 2 =2 3 Ans.[A] Hindu Method {Sri Dharacharya Method} (Discriminant Formula) : By completing the perfect square as b c x+ =0 a a ax2 + bx + c = 0  x2 + Adding and substracting LMF b I MNGH x  2a JK 2  b 2  4 ac Which gives, 4a2 FG b IJ 2 H 2a K OP PQ = 0 x = b  b2  4ac 2a 0) has two roots, given by = and In Quadratic equation ax2 + bx + c = 0, the term b2 – 4ac is called discriminant of the equation, which plays an important role in finding the nature of the roots. It is denoted by  or D. (A) Suppose a, b, c  R and a  0 then Hence the Quadratic equation ax2 + bx + c = 0 (a  b  b 2  4ac 2a NATURE OF ROOTS (i) If D > 0 Roots are Real and unequal (ii) If D = 0 Roots are Real and equal and each equal to –b/2a (iii)If D < 0 Roots are imaginary and unequal or complex conjugate. (B) Suppose a, b, c  Q, a  0 then b  b2  4ac  2a Note : Every quadratic equation has two and only two roots. (i) If D > 0 and D is perfect square Roots are unequal and Rational (ii) If D > 0 and D is not perfect square Roots are irrational and unequal CONJUGATE ROOTS Ex.1 The roots of the equation x2 – 2x – 8 = 0 are (A) – 4, 2 (C) 4, 2 Sol. (B) 4, – 2 (D) – 4,– 2 Quadratic Equation x2 – 2x – 8 = 0 The Irrational and complex roots of a quadratic equation are always occurs in pairs. Therefore (a, b, c,Q) If One Root then  + i  – i +  Other Root –  After factorization (x – 4) (x + 2) = 0 x = 4, – 2 Ans.[B] 28 Quadratic Equation C = 2 (a2 + b2) 1, B = –2 (a+b), Sol. B2 – 4AC = 1[2(a + b)]2 – 4(1) (2a2 + 2b2) Ex.3 The roots of the equation x  2 2x  1  0 are = 4a2 + 4b2 + 8ab – 8a2 – 8b2 (A) Imaginary and different = – 4a2 – 4b2 + 8 ab (B) Real and different = – 4(a–b)2 < 0 So roots are imaginary and different. Ans.[D] 2 (C) Real and equal (D) Rational and different Sol. The discriminant 2 2 2 of the equation Ex.6  4 1 1  8  4  4  0 and a perfect square, so roots are real and different but we can’t say that roots are rational because coefficients are not rational therefore. 2 2 ,   2 2 2 4 2 2 22   2  1 this 2 Sol. (a+b+c)2 – 4(b+c) (a) = a2+ b2+ c2 + 2ab + 2bc + 2ca – 4(b+c)a is irrational = a2 + b2 + c2 + 2ab + 2bc + 2ca – 4ab – 4 ac  The roots are real and different. = a2 + b2 + c2 – 2 ab + 2 bc – 2 ca Ans. [B] Ex.4 The roots of the equation (b+c) x2 – (a+b+c) x + a = 0 are (a,b,c  Q) (A) Real and different (B) Rational and different (C) Imaginary and different (D) Real and equal The discriminant of the equation is (a–b–c)2 > 0 If the roots of the equation x 2  2x  p  0 are So roots are rational and different. Ans.[B] real then the value of p is Sol. (A) p  1 (B) p  2 (C) p  3 (D) None of these Here a = 1, b = 2, c = p  discriminant  2 2 SUM AND PRODUCT OF ROOTS If  and  are the roots of quadratic equation ax2 + bx + c = 0, then, (i) Sum of Roots 4 1 p 0 (since roots are real) (ii) Product of Roots = 4 – 4p  0  4  4p P = =  p 1 e.g. Ans. [A] Ex.5 Coefficient of x b =– a Coefficient of x 2 S c a = cons tan t term coefficient of x 2 In equation 3x2 + 4x – 5 The roots of the quadratic equation x2 – 2 (a + b) x + 2(a2 + b2) = 0 are - Sum of roots S = (A) Rational and different (B) Rational and equal Product of roots P = (C) Irrational and different (D) Imaginary and different 1. =0 4 , 3 5 3 Relation between Roots and Coefficients If roots of quadratic equation ax2 + bx + c = 0 (a 0) are and then : 29 b (i) g= b g 2 b = 4 2 a 2 (ii) 2 = (iii) = 2 2 =– a b = 2 e = {b 4 (C) 2, 3/2 (D) 3/2, 1 b g= b a 2 3 j g (ix) a2 2 + e b b2 2 2 –2 2 –2 2 ac j = b 2 Ex.9 a2 2 Sol. 2 2  k k = 3, 3/2 Ans.[B] are roots of the equation ax2 + bx + c = 0 If , 1 2 2 Sol. 2 = 2 2 (a +b) = – c / 2 ...(1) Similarly 2 If the product of the roots of the quadratic equation mx2 – 2x + (2m–1) = 0 is 3 then the value of m is (B) 2 (D) 3 2m 1 Product of the roots c/a = 3 = m 3m – 2m = – 1 is -  (a +b) + c = 0 1 (A) 1 (C) –1 2 +b +c=0 (a +b) = – c / Ex.7 1 + 2 2 (B) ac ac 2 2 b2 (C) (D) 2 a2 c 2 a c Since , are the root of the ax2 + bx +c then a 4 2 3k – 6 = (A) g 4 =  2 4 ac 2 b = ) bk 2g then the value of b2 2 ( bk 2g c2 g– g g 2 2 j b = FG IJ + FG IJ H K H K 2 2 8 2 = a4 = 2 I JK 2 } je = (viii) 2 + 2 2 e (vii) 2 2 k 2 = 2 } 2 4 4 {b = 2 = k 2 , b a3 g bk 4g b g g = FGH k 42 IJK b Now g= – e g g (B) 3,3/2  3= F =G H (xi) D (A) 1,3 a3 (vi) (x) 2 3 g j Sol. = 3 = (v) 4 Ex.8 2 4 g –3 b g +3 b b =b (iv) D a = 2 a 2 = 4ac Quadratic Equation If the equation – (k–4) x –2 = 0 has difference of roots as 3 then the value of k is(k–2)x2 2 c 2 + ...(2) 1 2 c 2 c 2 1 b c/ g 2 + + 2 2 = c = 2 2 = = = 2 2 2 a c c b 1 c/ g 2 2 2 Ans.[C] m=–1 Ans. [C] 30 FORMATION OF AN EQUATION Quadratic Equation Some examples of symmetric expressions are– WITH GIVEN ROOTS (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) A quadratic equation whose roots are given by  and  is FG IJ FG IJ H K H K 2 x   x   0  x 2  x   x    0 2 2  x     x    0 i.e., x2  (Sum of roots)x + Product of roots = 0 1. Transformation of an Eqaution Ex.10 If ,  are the root of a quadratic equation x 2  3x  5  0 then the equation whose roots are If , are roots of the equation ax 2  bx  c  0 then the equation whose roots are (Replace x by (i) 1 ) x (Replace x by (ii)  2  3  7 and 2  3  7 is (A) x 2  4x  1  0 (C) x 2  4x  4  0 Sol. x ) (B) x2  4x  1  0 (D) x 2  2x  3  0 Since , are the roots of equation x 2  3x  5  0 So 2  3  5  0 ; 2  3  5  0 (iii)   2  3  5 ..... (i) {Replace x by x  k }  2  3   5 (iv) ..... (ii) Putting the value from (i) and (ii) in  2  3  7 (Replace x by x ) 1/ n and 2  3  7 (v) We get 5  7 and 5  7 (Replace x by xn )  2 and 2 are the roots (vi)  The required equation is x 2  4x  4  0 (Replace x by x ) k Ans.[C] Ex.11 The quadratic equation whose one root is (vii) 2 5 will be k k (A) x 2  4x  1  0 (C) x 2  4x  1  0 (Replace x by kx) 2. 1 Symmetric Expressions The symmetric expressions of the roots , of an equation are those expressions in and , which do not change by interchanging and . To find the value of such an expression, we generally express that in terms of and . Sol. Given root  1 2 5 (B) x2  4x  1  0 (D) None of these  5 2 So the other root   5  2 . Then sum of the roots  4 , product of the roots  1 Hence the equation is x2  4x  1  0 Ans.[A] 31 Ex.12 The equation whose roots are 3 and 4 will be(A) x2 + 7x + 12 = 0 (B) x2 – 7x + 12 = 0 (A) x2 + x –2 = 0 (B) x2 + 2x – 8= 0 (C) x2 – x + 12 = 0 (C) x2 –x –2 = 0 (D) None of these (D) x2 + 7x – 12 = 0 Sol. Quadratic Equation Ex.15 If , are roots of the equation 2x2 + x –1 =0 then the equation whose roots are 1/ , 1 / will be - From the given equation Sol. The quadratic equation is given by = 1/ 2 , x2 – (sum of the roots) x + (product of roots) =0 The required equation is- The required equation = x2 – (3+4) x x2 – + 3.4 = 0 = x2 – 7x + 12 = 0 Sol. (C) x2 + 4x – 1= 0 (D) x2 + 2x + 1=0 IJ x + K x2 – Ex.13 The quadratic equation with rational coefficients whose one root is 2 + 3 is (B) x2 + 4x + 1=0 FG H FG IJ x + H K 1 / 2I x2 – FG H 1 / 2JK x + Ans. [B] (A) x2 – 4x + 1= 0 = 1/ 2 1 1 =0 =0 1 =0 1/ 2 x2 –x–2 = 0 Ans.[C] Short cut : Replace x by 1/x The required equation is 2(1/x)2 + 1/x – 1 = 0 x2 –x–2 = 0 x2 – {(2+ 3 ) + (2– 3 ) } x ROOTS UNDER PARTICULAR CASES + (2 + 3 ) (2– 3 ) = 0 x2 – 4x + 1 = 0 or Ans. [A] (i) Ex.14 If , are roots of the equation x2 – 5x + 6 = 0 then the equation whose roots 3 and are 3 is- (ii) (A) x2 – 11x + 30 = 0 (B) (x–3)2 – 5 (x–3) + 6 = 0 (iii) (iv) (C) Both (1) and (2) For the quadratic equation ax2 + bx + c = 0 If b = 0  roots are of equal magnitude but of opposite sign If c = 0  one root is zero other is – b/a If b = c = 0  both root are zero If a = c  roots are reciprocal to each other (D) None Sol. Let a0 c0 (v) If a  0 c  0 3=x UV W  signs = x – 3 (Replace x by x – 3) So the required equation is (x – 3)2 – 5 (x–3) + 6 = 0 ...(1) (vi) If  x2 – 6 x + 9 –5x + 15 + 6 = 0 x2 –11 x + 30 = 0 ...(2) Ans.[C] If Roots are of opposite (vii) If 0, 0, 0, 0, 0, 0, 0, 0, UV  W 0U V 0W 0 0 both roots are negative. both roots are positive. (viii) If sign of a = sign of b sign of c Greater root in magnitude is negative. 32 Quadratic Equation (ix) If sign of b = sign of c sign of a Ex.19 The real values of a for which the quadratic Greater root in magnitude is positive. 2x 2  a3  8a  1 x  a 2  4a  0 equation (x) If a+b+c = 0 possesses roots of opposite signs are given by one root is 1 and second root is c/a. (A) a  5 (B) 0  a  4 (xi) If a = b = c = 0 then equation will become an identity (C) a  0 (D) a  7 and will be satisfy by every value of x. Sol. The roots of the given equation will be of opposite signs if they are real and their product is negative, 2 i.e., Discriminant  0 and product of roots < 0. Ex.16 The roots of the equation x –3x –4 = 0 are– Sol. (A) Opposite and greater root in magnitude is positive (B) Opposite and greater root in magnitude is negative (C) Reciprocal to each other (D) None of these The roots of the equation x2 –3x –4 = 0 are of opposite sign and greater root is positive ( a >0 , b< 0, c < 0) Ans.[A] 2  a3  8a  1  8 a2  4a  0 and a2  4a 0 2  a 2  4a  0  a2  4a  0  a3  8a  1 2  8 a 2  4a  0    Ans.[B] 0a4 CONDITION FOR COMMON ROOTS 1. Only One Root is Common : Let  be the common 2 root of quadratic equations a1x2 + b1x + c1 = 0 and Ex.17 The roots of the equation 2x – 3x + 2 = 0 are a2x2 + b2x + c2 = 0 then (A) Negative of each other (B) Reciprocal to each other  a12 + b1 + c1 = 0 (C) Both roots are zero a22 + b2 + c2 = 0 (D) None of these By Cramer's rule : Sol. The roots of the equations 2x2 –3x + 2 = 0 are 2  1   reciprocal to each other because here a = c  c1 b1 a1  c 1 a1 b1 Ans.[B] c 2 b2 a2  c 2 a 2 b2 2 k or 1 Ex.18 If equation = has equal and k 1 opposite roots then the value of k is (A) (C) Sol. (B) a b 1 (D) a a b   1   b1c 2  b2 c1 a 2 c1  a1c 2 a1b 2  a2b1 b  a b 1 Let the roots are & – . given equation is (x2 – bx) (k+1) = (k–1) (ax–c) x2 (k+1) – bx(k+1) = ax (k–1) –c (k–1) x2 (k+1) –bx (k–1) – ax (k–1) + c (k–1) = 0 Now sum of roots = 0 (  – = 0) b (k+1) + a (k–1) = 0 k= 2 Ans.[B] a 2 c1  a1c 2 2 b1c 2  b 2 c1 ,   0. ,  a1b2  a2b1 a1b2  a2b1  The condition for only one Root common is (c1a2 – c2a1)2 = (b1c2 – b2c1)(a1b2 – a2b1) 2. Both roots are common : Then required conditions is a1 b c = 1 = 1 a2 b2 c2 Note: Two different quadratic equation with rational coefficient cannot have single common root which is complex or irrational, as imaginary and surd roots always occur in pair. 33 Quadratic Equation 2 FG 2 IJ H 3 K Ex.20 If one root of the equations x2+ 2x + 3k = 0 and 2x2 + 3x+ 5k = 0 is common then the values of k is (A) 1, 2 (C) 1, 3 Sol. (B) 0, –1 (D) None of these = 2 2 1  = 6k  5k = –k ...(1) = –k ...(2) 3 2 , 3 1 Ans.[C] 3 Sol. (A) k 2  7k  1  0 (B) k 2  7k  1  0 (C) k 2  7k  1  0 (D) k 2  7k  1  0 Let the common root is  then 2    1  0 2 2    k  0 = k2 k2 = – k 2 1 Ex. 23 If x 2  x  1  0 and 2x 2  x  k  0 have a common root then Since one root is common, let the root is . 2 1 3 By cross multiplication k2 + k = 0 k (k + 1) = 0 k = 0 and k = –1 Ans. [B] Ex.21 If the equations 2x 2 + x + k = 0 and x2 + x/2 –1 = 0 have 2 common roots then the value of k is- 2  1   k  1 2  k 1  2 2  k  1 1 k 2k  ,  3 3 3 2 Sol. (A) 1 (B) 3 (C) –1 (D) –2 1 k 2k 2  3   3  k  7k  1  0   Since the given equation have two roots in common so from the condition a1 a2 = b1 b2 2 1 = 1 1/ 2 k =–2 k 1 Ans.[D] Ex.22 If x2 + x – 1 = 0 and 2x2 – x +  = 0 have a common root then – (A)  2 – 7 + 1 = 0 2 (C)  + 7 + 1 = 0 Sol. 2 (B) + 7 – 1 = 0 (D) 2 – 7 – 1 = 0 Let the common root is  then QUADRATIC EXPRESSION The expression ax2 + bx + c is said to be a real quadratic expression in x. Where a, b and c are real a  0. Let y = ax2 + bx + c c1 = c 2 = Ans. [B]  y= a 2  b  D  x    2  2a  4a    D  b      2   y  4a   a  x  2a  .....(1) where D = b2 – 4ac 2 +  – 1 = 0 Equation (1) represents a parabola with vertex at 22 –  +  = 0 D   b A ,  , and axis of this parabola is parallel 2 a 4 a  By cross multiplication 2 1 to y axis and it is x = – b/2a. 34 GREATEST AND LEAST VALUE OF QUADRATIC EXPRESSION (i) If a  0 , then the quadratic expression y  ax 2  bx  c has no greatest value but it has least value 4ac  b2 b at x   4a 2a Quadratic Equation Ex.26 The minimum value of the expression 4x2 + 2x + 1 is(A) 1/4 (C) 3/4 Sol. (B) 1/2 (D) 1 Since a = 4 > 0 therefore its minimum value is = (ii) If a  0 , then the quadratic expression y  ax 2  bx  c has no least value but it has greatest Ex.24 The range of the values of x for all real value x 4 2 (A) 10 (C) 25 1 1 y 2 2 (B) 1  y  1 (C) 1 1 y 6 6 (D) None of these 4 Sol. 12 3 = 16 4 Since a = –4 < 0 therefore its maximum value is = = 80 400 16 4 = 480 = 30 16 Ans.[D] FEW GRAPHS OF QUADRATIC x 2 x 4 EXPRESSION (i) y = x2 (a = 1, b = 0, c = 0)  x 2 y  x  4y  0 Now, x  R  B 2  4AC  0  1  4y.4y  0 Vertex (0, 0) Axis of the parabola x = 0  4y  1 4y  1  0 This parabola opens in upward direction. 1 1  y 4 4 Ans.[B] (ii) y = – x2 + 2x + 1 (a = – 1, b = 2, c = 1) Ex.25 If the roots of the quadratic equation x 2  4x  log3 a  0 are real, then the least value of a is (B) 1 81 (C) 64 (D) None of these Since the roots of the given equation are real.  log3 a  4  a  3 4  a  vertex (1,2) Axis of the parabola x = 1 This parabola opens downward. Note: If a > 0, shape of parabola is upward. 1 Sol. = (B) 20 (D) 30 2 (A) (A) 81 16 Ans.[C] of x is Let y  = Ex.27 The maximum value of 5 + 20 x – 4x2 for all real value of x is- 4ac  b2 b at x   value 4a 2a Sol. 4( 4)(1) ( 2) 2 4( 4 ) 1 81 1 Hence, the least value of a is 81 Ans.[B] And if a < 0, shape of parabola is downward. Values of x where curve crosses x-axis will be roots of the equation ax2 + bx + c = 0 as y = 0 on these points. If a curve is above x-axis for all x, then it means that it does not intersect x-axis or we can say equation ax2 + bx + c = 0 have imaginary roots. This gives rise to following cases: 35 Quadratic Equation Cases: LOCATION OF ROOTS Let f(x) = ax2 + bx + c where a, b, c,  R (a  0). (Interval in which roots lie) In some problems we want the roots of the equation For some value of x, f(x) may be positive, negative ax 2  bx  c  0 to lie in a given interval. For this we or zero. This gives rise to the following cases: impose conditions on a, b and c. Since a  0 , we (i) a > 0 and b2 – 4ac < 0  f(x) > 0 ,  x  R can take f x  x 2  b x  c . a a In this case whole of the parabola lies above the x axis. (i) If both the roots are positive i.e., they lie in 0, , (ii) a > 0 and b2 – 4ac = 0  f(x)  0,  x  R In this case whole of the parabola lies above the xaxis except at one point where it touches the x – axis. The point is . then the sum of the roots as well as the product of the roots must be positive.      b  0 and   c  0 a a with b2  4ac  0 . Similarly, if both the roots are negative i.e. they lie in ,0 then the sum of the roots must be negative 2 (iii) a > 0 and b – 4ac > 0 and the product of the roots must be positive. Let f(x) = 0 have two real roots  and  i.e.      b  0 and   c  0 with b2  4ac  0 (say  < ) a then f(x) > 0,  x  (–  , )  (,  ) and f(x) < 0,  x  (, ) (iv) a < 0 and b2 – 4ac < 0  f(x) < 0,  x  R a Both the roots are of the same sign if a and c are of same sign. Now if b has the same sign that of a, both roots are negative or else both roots are positive. If a and c are of opposite sign both roots are of opposite sign. In this case whole of the parabola lies below the x– axis. (ii) Both the roots are greater than given number k if the following three conditions are satisfied (v) a < 0 and b2 – 4ac = 0  f(x)  0,  x  R and f k  0 . In this case whole of the parabola lies below the x– axis, except at one point where it touches the x–axis. The point is . (vi) a < 0 and b2 – 4ac > 0 Let f(x) = 0 have two real roots  and  ( < ). Then f(x) < 0,  x  (–  , )  (,  ) and f(x) > 0,  x  (, ) 36 Quadratic Equation (iii) Both the roots will be less than a given number k if (vi) A given number k will lie between the roots if the following conditions are satisfied: f k  0 , D > 0. and f k  0 In particular, the roots of the equation will be of (iv) Both the roots will lie in the given interval k1,k 2 if opposite signs if 0 lies between the roots  f 0  0 . the following conditions are satisfied: and Ex.28 If f(x) is a quadratic expression which is positive for all real values of x and g(x) = f(x) + f '(x) + f "(x) , then for any real value of x(A) g (x) < 0 (C) g (x) = 0 Sol. (B) g (x) > 0 (D) g (x) 0 Let f(x) = ax2 + bx + c, then g(x) = (ax2 + bx + c) + 2ax + b + 2a = ax2 + (b + 2a) x + (c + b + 2a) (v) Exactly one of the root lies in the given interval k1,k 2 2  f(x) > 0 , therefore b – 4ac < 0 and a > 0. if f k1 .f k 2  0 Now for g(x), Discriminant = (b+2a)2 – 4a (c+b+ 2a) = b2 – 4ac– 4a2 < 0 ( b2 – 4ac < 0, – 4a2 < 0) Therefore signs of g(x) and a are same i.e. g(x) > 0. Ans.[B] Ex.29 For real values of x, 2x2 + 5x – 3 > 0, if(A) x < – 2 (C) x > 1 Sol. (B) x > 0 (D) None of these Discriminant b2 – 4ac = 25 + 24 = 49 > 0 Roots are real. The given expression is positive for those real values of x for which x ( – 3, 1/2), because a = 2 > 0. x > 1 is true. Ans.[C] 37 Quadratic Equation THEORY OF EQUATIONS (i) If p x  anxn + an-1xn-1 + ... + a0 ,(a0, a1, a2, ........,an  R and an  0) then p(x) = 0 has exactly n roots. (real/complex) Some Important Points (i) Every equation of nth degree n  1 has exactly n roots and if the equation has more than n roots, it is an identity. (ii) Imaginary roots always occur in conjugate pairs. If (ii) If roots of quadratic equations a1x 2  b1x  c1  0   0 and  + i is a root of p(x), then  – i is and a2 x 2  b2 x  c 2  0 are in the same ratio also a root. (iii) A polynomial equation in x of odd degree has at least one real root (moreover it has odd no. of real roots). (iv) If x1,.....,xn are the roots of p(x) = 0, then p(x) can be written in the form p(x)  an(x – x1)...(x – xn). (v) If  is a root of p(x) = 0, then (x – ) is a factor of p(x) and vice - versa. (vi) If x1,....,xn are the roots of p(x)  anxn + ... + a0 = 0, an  0. n Then  x i   i 1 a n 1 , an a  x i x j  n 2 , 1i  j n an  1  2    i.e.  then b22 1 2   (iii) If one root is k times the other root of quadratic 2 equation a1x 2  b1x  c1  0 then QUADRATIC EXPRESSION IN TWO VARIABLE The general form of a quadratic expression in two variables x, y is ax2 + 2hxy + by2 + 2gx + 2fy + c The condition that this expression may be resolved into two linear rational factors is or ao x1 x2...xn = (–1)n an . (vii) If equation f x  anxn + an–1xn–1 + ... + a1x + a0 = 0 has more than n distinct roots then f(x) is identically zero. (viii) If p(a) and p(b) (a < b) are of opposite sign, then p(x) = 0 has odd number of real roots in (a, b), i.e. it has at least one real root in (a, b) and if p(a) and p(b) are of same sign then p(x) = 0 has even number of real roots in (a, b). (ix) If coefficients of p(x) (polynomial in x written in descending order) have ‘m’ changes in signs, then p(x) = 0 have at the most ‘m’ positive real roots and if p(– x) have ‘t’ changes in sign, then p(x) = 0 have at most ‘t’ negative real roots. By this we can find maximum number of real roots. ah g g f c Example : For what value of m the expression y2 + 2xy + 2x + my – 3 can be resolved into two rational factors? Sol. Here a = 0, b = 1, c = – 3 h = 1, g = 1, g = m/2 0 1 1 So   0  1 1 m / 2 0 1 m/2 3  (m/2 + 3) + (m/2 – 1) = 0  m+2=0  m=–2 38 Quadratic Equation Ex.30 For what value of m the expression y2 + 4xy + 4x + my – 2 can be resolved into two rational factors(A)1 (C) 2 Sol. (B) –1 (D) –2 Here a = 0 , b = 1, c = –2 So =0 0 2 2 1 m/ 2 m/2 =0 2(4+m) + 2 (m–2) = 0 4m+4=0 m = –1 Ans. [B] h = 2 , g = 2, f = m/2 39