PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/1kElG06RhBRuPvr_UWgdIBRfq59YsFEv6/edit?usp=share_link&ouid=109474854956598892099&rtpof=true&sd=true Quadratic Equations Quadratic Equations in Real and Complex Number System and their Solutions. Relation between Roots and Co-efficients, Nature of Roots, Formation of Quadratic Equations with given Roots. C H A P T E R QUADRATIC EQUATIONS WITH REAL COEFFICIENTS The expression, ax2 + bx + c, where a, b, c ∈ R and a ≠ 0 is called as a quadratic expression in x. The quadratic expression when equated to zero, ax2 + bx + c = 0, is called as a quadratic equation in x. Where the numbers a, b, c are called the coefficients of the equation. The values of x which satisfy the quadratic equation is called roots (also called solutions or zeros) of the quadratic equation. This equation has two roots which are given by x = − b ± D , 2a where D (or Δ) = b2 – 4ac is called as discriminant of the quadratic equation. If α and β denote the roots of ax2 + bx + c = 0, then, CHAPTER INCLUDES : Quadratic Equations with Real coefficients Nature of Roots Relation between Roots and Coefficients Common Roots Formation of Quadratic Equation with given Roots (i) (ii) α + β = − b a αβ = c a Symmetric Functions of Roots Quadratic Expressions (iii) α − β = D | a | Theory of Polynomial Equations Wavy Curve Method If α and β are the roots of ax2 + bx + c = 0 and γ, δ are those of lx2 + mx + n = 0, find the equation whose roots are αγ + βδ and αδ + βγ. Solution : Equations Reducible to Quadratic Equations General idea about ax 2 + bx + c = 0 has roots α and β b c Miscellaneous Equations ∴ α + β = − ; a αβ = a Important Theorems lx 2 + mx + n = 0 has roots γ and δ and Results Solved Examples ∴ γ + δ = − m ; l γ δ = n l The required quadratic equation is x 2 − (αγ + βδ + αδ + βγ)x + (αγ + βδ)(αδ + βγ) = 0 where αγ + βδ + αδ + βγ = (α + β)(γ + δ) = b . m a l (αγ + βδ)(αδ + βγ) = (α2 + β2 )γδ + αβ(γ2 + δ2 ) n ⎛ b2 − 2ac ⎞ c ⎛ m2 − 2n l ⎞  l n (b2 − 2ac) + ac(m 2 − 2n l ) ⎜ ⎟ ⎜ ⎟ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ a2l 2 ∴ a2l 2 ⋅ x 2 − ablm x + l n (b2 − 2ac) + ac(m2 − 2n l ) = 0 Identity If a quadratic equation is satisfied by three different values of variable (x) i.e. x = x1, x2, x3, then it is an identity and it will be satisfied by all values of x. Identical Equations Two equations having same roots are called identical equations. NATURE OF ROOTS (i) D > 0 ⇔ roots are real and unequal. D = 0 roots are real and equal to − b 2a D < 0 ⇔ roots are complex and unequal. If D is a perfect square of a rational number, then the roots are rational and unequal. If D > 0 and is not a perfect square then the roots are irrational and unequal. (i) Imaginary roots always occur in pair if a, b, c are real numbers. i.e., if α + iβ is one root of quadratic equation, then α − iβ will be its other root. Irrational roots of a quadratic equation with rational coefficients always occur in pair. i.e., if m + is one root of equation, then its other root will be m − n . Determine the values of m for which the equation 5x 2 − 4x + 2 + m(4x 2 − 2x − 1) = 0 equal roots product of the roots as 2 Solution : 5x 2 − 4x + 2 + m(4x 2 − 2x − 1) = 0 will have (5 + 4m)x 2 − 2(m + 2)x + (2 − m) = 0 …(1) (a) For equal roots, the discriminant of (1) should be equal to zero. ∴ 4(m + 2)2 − 4.(5 + 4m)(2 − m) = 0 m2 + 4m + 4 − (10 + 8m − 5m − 4m2 ) = 0 5m2 + m − 6 = 0 (5m + 6) (m – 1) = 0 m = − 6 5 or m = 1 (b) 2 − m = 2 5 + 4m ∴ 2 – m = 10 + 8m m = − 8 9 RELATION BETWEEN ROOTS AND CO-EFFICIENTS If α and β are the roots of ax2 + bx + c = 0 then, ax2 + bx + c = a(x – α)(x – β) = a(x2 –(α + β)x + αβ) If Δ = b2 – 4ac > 0. Now consider the following cases. Nature of roots Case-I a > 0, b > 0, c > 0 ⇒ α + β < 0, αβ > 0 Both roots are negative. Case-II a > 0, b > 0, c < 0 ⇒ α + β < 0, αβ < 0 Both roots are opposite in sign; Magnitude of negative root is more than the magnitude of positive root. Case-III a > 0, b < 0, c > 0 ⇒ α + β > 0, αβ > 0 Both roots are positive. Case-IV a > 0, b < 0, c < 0 ⇒ α + β > 0, αβ < 0 Roots are opposite in sign. Magnitude of positive root is more than magnitude of negative root. Condition for a root of a quadratic equation ax2 + bx + c = 0 to be reciprocal of the root of a´x2 + b´x + c´ = 0 is (cc′ − aa′)2 = (ba′ − b′c)(ab′ − bc′) COMMON ROOTS One root common : If α ≠ 0 is a common root of the equation a1x2 + b1x + c1 = 0 ...(i) and a2x2 + b2x + c2 = 0 ...(ii) then we have a1α2 + b1α + c1 = 0 and a2α2 + b2α + c2 = 0 These give α2 b c − b c = α c a − c a = 1 a b − a b (a1b2 − a2b1 ≠ 0). 1 2 2 1 1 2 2 1 1 2 2 1 Thus, the required condition for one common root is (a1b2 – a2b1) (b1c2 – b2c1) = (c1a2 – c2a1)2 and the value of the common root is α = c1a2 − c2a1 a1b2 − a2b1 Both roots common : or b1c2 − b2c1 . c1a2 − c2a1 If the equations (i) and (ii) have both roots common, then these equations will be identical. Thus the required condition for both roots common is a1 = b1 = c1 (If no root is equal to zero) a2 b2 c2 Find m and n in order that the equations mx2 + 5x + 2 = 0 and 3x2 + 10x + n = 0 may have both the roots common. Solution : The equations are mx2 + 5x + 2 = 0 and 3x2 + 10x + n = 0. Since they have both the roots common, m = 5 = 2 3 10 n Θ a1 = b1 = c1 a2 b2 c1 From the first-relation, m = 15 = 3 . 10 2 From the last-relation, n = 4. FORMATION OF EQUATION WITH GIVEN ROOTS If α and β are the roots of ax2 + bx + c = 0, a ≠ 0 then ax2 + bx + c ≡ a(x – α) (x – β) ⇒ x 2 + b x + c ≡ x2 – (α + β) x + αβ a a Hence, x2 – (sum of roots) x + product of roots = 0 is the required quadratic equation. SYMMETRIC FUNCTION OF THE ROOTS A function of α and β is said to be a symmetric function if it remains unchanged when α and β are interchanged. For example, α2 + β2 + 2αβ is a symmetric function of α and β whereas α2 – β2 + 3αβ is not a symmetric function of α and β. In order to find the value of a symmetric function of α and β, express the given function in terms of α + β and αβ. The following results may be useful. (i) α2 + β2 = (α + β)2 – 2αβ (ii) α3 + β3 = (α + β)3 – 3αβ (α + β) (iii) α4 + β4 = (α3 + β3) (α + β) – αβ (α2 + β2) (iv) α5 + β5 = (α3 + β3) (α2 + β2) – α2β2 (α + β) (v) α −β = (vi) α2 – β2 = (α + β) (α – β) (vii) α3 – β3 = (α – β) [(α + β)2 – (viii) α4 – β4 = (α + β) (α – β) (α2 QUADRATIC EXPRESSION Let y = ax2 + bx + c ⎛ = a ⎜⎜ x + ⎝ b ⎞2 ⎟ – 2a ⎠ D ⎟ 4a2 ⎟ ⎛ – b , – D ⎞ …(1) –b represents a parabola with vertex ⎜ 2a ⎟ 4a ⎠ and axis of the parabola is x = 2a If a > 0, the parabola opens upward while if a < 0, the parabola opens downward. The parabola cuts the x-axis at points corresponding to roots of ax2 + bx + c = 0. If this equation has D > 0, the parabola cuts x-axis at two real and distinct points. D = 0, the parabola touches x-axis at D < 0, then; x = –b . 2a if a > 0, parabola lies above x-axis. if a < 0, parabola lies below x-axis. 1. y = x2 = x2 2. y = – x2 + 3x – 2 y = –x2+3x–2 Graphs of Quadratic Expressions Let f(x) = ax2 + bx + c and α, β : α < β , be its roots a > 0 and D < 0 ⇔ f (x) > 0 ∀ x ∈ R α and β are complex conjugates b 2a a > 0 and D = 0 ⇔ f (x) ≥ 0 ∀ x ∈ R f(x) = 0 at x = –b 2a b 2a a > 0 and D > 0 ; then, f(x) > 0 ∀ x ∈ (–∞, α) ∪ (β, ∞) f(x) < 0 ∀ x ∈ (α, β) f (x) min = − D 4a at x = − b 2a b a < 0 and D < 0 ⇔ f (x) < 0 ∀ x ∈ R 2a α and β are complex conjugates a < 0 and D = 0 ⇔ f (x) ≤ 0 ∀ b x ∈ R 2a f(x) = 0 at x = –b 2a a < 0 and D > 0 ; then f (x) < 0 ∀ x ∈(−∞, α) ∪ (β, ∞) a b b f (x) > 0 ∀ x ∈(α, β) 2a f (x ) max = − D 4a at x = − b 2a Location of Roots Let f (x) = ax 2 + bx + c, a,b,c ∈ R, a ≠ 0 and α,β be roots of f (x) = 0 A real number k lies between the roots of f (x) = 0 ⎛ – b , – D ⎞ ⎜ ⎝ 2a ⎟ 4a ⎠ D > 0 af (k) < 0, where α < β ⎛ – b ⎜ , ⎝ 2a –D ⎞ ⎟ 4a ⎠ If both roots of quadratic equation f (x) = 0 are greater than k, then a < 0 ⎛ − b ⎜ − D ⎞ ⎟ ⎝ 2 a 4 a ⎠ D ≥ 0 af (k) > 0 ⎛ − b ⎜ , ⎝ 2 a − D ⎞ ⎟ 4 a ⎠ (iii) k < −b , where α ≤ β 2a If both roots are less than real number k, then a > 0 α β k x-axis a < 0 ⎛ − b ⎜ , ⎝ 2a − D ⎞ ⎟ 4a ⎠ D ≥ 0 af (k) > 0 ⎛ − b − D ⎞ ⎜ , ⎟ k α β x-axis (iii) k > −b , where α ≤ β 2a ⎝ 2a 4a ⎠ Exactly one root lies between real numbers k1 and k2, where k1 < k2. ⎛ − b , − D ⎞ ⎜ ⎝ 2 a ⎟ 4 a ⎠ D > 0 f (k1).f (k2) < 0, where α < β If both roots of f (x) = 0 are confined between real numbers k1 and k2, where k1 < k2. a < 0 ⎛ − b − D ⎞ , ⎜ ⎝ 2a ⎟ 4a ⎠ D ≥ 0 af (k1) > 0 af (k2) > 0 ⎛ − b ⎜ , ⎝ 2a − D ⎞ ⎟ 4a ⎠ (iv) k < −b < k 1 2a 2 , where α ≤ β If real numbers k1 and k2 lie between roots of f (x) = 0, where k1 < k2. D > 0 af (k1) < 0 af (k2) < 0, where α < β NUMBER OF ROOTS OF A POLYNOMIAL EQUATION If f (x) is an increasing function in [a, b], then f (x) = 0 will have atmost one root in [a, b]. Let f (x) = 0 be a polynomial equation and a, b are two real numbers. Then f (x) = 0 will have at least one real root or odd number of real roots in (a, b) if f (a) and f (b) (a < b) are of opposite signs. A(a,f (a)) A(a,f (a)) (a, 0) P (b, 0) (a, 0) P R S Q T (b, 0) B(b,f (b)) One real root Odd number of real roots B(b,f (b)) But if f (a) and f (b) are of same signs, then either f (x) = 0 has no real root or even number of real roots in (a, b). A(a, f (a)) B(b, f (b)) (a, 0) (b, 0) No real root Even number of real roots If the equation f (x) = 0 has two real roots a and b then f ′(x) = 0 will have atleast one real root lying between a and b (using Rolle’s Theorem). WAVY CURVE METHOD Let (x − a1)k1 (x − a2 )k2 (x − an )kn F (x) = (x − b )r1 (x − b )r2 ..........(x − b )rn 1 2 n where k1, k2,...........kn & r1, r2,...........rn ∈ N and a1, a2,...........an & b1, b2, bn are fixed real numbers. Points where numerator becomes zero are called zero’s or roots of the function and where denominator becomes zero are called poles of the function. Find poles and zeros of the function F(x). The corresponding zeros are a1, a2,...........an and poles are b1, b2, bn. Mark the poles and zeros on the real number line. If there are n poles & zeros the entire number line is divided into ‘n + 1’ intervals. For F(x), number line is divided into ‘2n + 1’ intervals. Place a positive sign in the right-most interval and then alternate the sign in the neighbouring interval if the pole or zero dividing the two interval has appeared odd number of times. If the pole or zero dividing the interval appeared even number of times then retain the sign in the neighbouring interval. The solution of F(x) > 0 is the union of all intervals in which plus sign is placed and the solution of F(x)<0 is the union of all intervals in which minus sign is placed. Let f (x) = (x − 6)100 (x + 4)441(x − 3)7(x + 2)16 11 x 200 ⎛ x − 1 ⎞ (x + 1)43 ⎜ ⎟ ⎝ 3 ⎠ 1 The critical points are – 4, – 2, – 1, 0, 3 , 3, 6 Solution : f (x) > 0 ∀ x ∈ (– ∞, – 4) ∪ (– 1, 0) ∪ ⎛ 0, 1 ⎞ ∪ (3, 6) ∪ (6, ∞) and ⎜ ⎟ ⎝ 3 ⎠ f (x) < 0 x ∈ (– 4, –2) ∪ (– 2, –1) ∪ ⎛ 1 ,3 ⎞ ∀ ⎜ ⎟ ⎝ 3 ⎠ EQUATIONS REDUCIBLE TO QUADRATIC EQUATIONS Reciprocal equation of the standard form can be reduced to an equation of half its dimension. Solve 2x4 + x3 – 11x2 + x + 2 = 0 Solution : Since x = 0 is not a solution hence, divide by x2 We get, 2x 2 + x − 11+ 1 + 2 = 0 x x 2 or ⎛ 2 1 ⎞ ⎛ 1 ⎞ 2 ⎜ x + ⎟+ ⎜ x + ⎟ − 11 = 0 x 2 x Let ⎝ ⎠ ⎝ ⎠ x + 1 = y x 2 (y2 – 2) + y – 11 = 0 or 2y2 + y – 15 = 0 ⇒ y = −3, 5 2 Corresponding values of x are 1 ,2, − 3 ± 5 2 2 Equations of the form (x – a) (x – b) (x – c) (x – d) = A where a < b < c < d can be solved by change of variable. y = x − ⎛ a + b + c + d ⎞ ⎜ ⎟ ⎝ 4 ⎠ (x – a) (x – b) (x – c) (x – d) = Ax2 where ab = cd can be solved by assumption y = x + ab . x (x + 4) (x + 6) (x + 8) (x + 12) = 1680x2 Here (– 4 × –12) = 48 = (–6) × (–8) (x + 4) (x + 12) (x + 6) (x + 8) = 1680x2 ⇒ (x2 + 16x + 48) (x2 + 14x + 48) = 1680x2 ⎛ x + 16 + 48 ⎞ ⎛ x + 14 + 48 ⎞ = 1680 ⇒ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ x ⎠ (Θ x ≠ 0) Let y = x + 48 x ⇒ (y + 16) (y + 14) = 1680 ⇒ y2 + 30y – 1680 + 224 = 0 ⇒ y2 + 30y – 1456 = 0 ⇒ y2 + 56y – 26y – 1456 = 0 ⇒ (y + 56) (y – 26) = 0 ⇒ ∴ y = 26, – 56 x + 48 = 26 x and x + 48 = −56 x ⇒ x2 – 26x + 48 = 0 ⇒ x2 + 56x + 48 = 0 ⇒ x = 2, 24 ⇒ x = = −28 ± ∴ x = {2,24,− 28 − 4 46,− 28 + 4 46 } = −28 ± 4 For the equation of the type (x – a)4 + (x – b)4 = A Substitute y = (x − a) +(x − b) 2 Important Theorems and Results Rolle’s Theorem – Let f (x) be a function defined on [a, b] such that f (x) is continuous on [a, b] f (x) is differentiable on (a, b) and f (a) = f (b) Then there exists a c ∈ (a, b) such that f ′(c) = 0 Lagrange’s theorem : Let f (x) be a function defined on [a, b] such that f (x) is continuous on [a, b] and f (x) is derivable on (a, b). Then c ∈ (a, b) such that f ′(c) = f (b) − f (a) b − a Factor theorem : If α is a root of the equation f (x) = 0, then f (x) is exactly divisible by (x – α) and conversely, if f (x) is exactly divisible by (x – α) then α is a root of the equation f (x) = 0. Every equation of an odd degree has atleast one real root, whose sign is opposite to that of its last term, provided that the coefficient of the first term is positive. Every equation of an even degree whose last term is negative has at least two real roots, one positive and one negative, provided that the coefficient of the first term is positive. If an equation has no odd powers of x, then all roots of the equation are complex provided all the coefficients of the equation are having positive sign. If x = α is root repeated m times in f (x) = 0, (f (x) = 0 is an nth degree equation in x) then f (x)= (x – α)m g (x) when g (x) is a polynomial of degree (n – m) and the root x = α is repeated (m – 1) times in f ′ (x) = 0, (m – 2) times in f ″ (x) = 0,...(m – (m – 1)) times in f m –1(x) = 0. The condition that a quadratic function f (x, y) = ax2 + 2hxy + by2 + 2gx + 2fy + c may be resolved into two linear factor if a h g Δ = abc + 2fgh – af 2 – bg 2 – ch2 or h b f g f c is 0. Law of proportions If a = c = e = ..., then each of these ratios is also equal to : b d f (i) a + c + e + ... b + d + f + ... ⎛ pan + qcn + ren + ... ⎞1/ n (ii) (iii) ⎜ ⎜ pbn qd n = rf n ⎟ + ... ⎟ (when p, q, r, n ∈ R) SOLVED EXAMPLES Example 1 : If the equations 4x2–11x + 2k = 0 and x2 – 3x – k = 0 have a common root, then the value of k and common root is (1) 0, 17 (2) −17 , 17 6 36 6 (3) 0, −17 6 (4) −17 , 0 6 Solution : The condition for the equations ax 2 + bx + c = 0 and a′x 2 + b′x + c′ = 0 to have a common root α is (ca′ − c′a)2 = (bc′ − b′c)(ab′ − a′b) 4x 2 − 11x + 2k = 0 x 2 − 3x − k = 0 and α = ca′ − c′a ab′ − a′b ∴ (2k + 4k )2 = (11k + 6k )(−12 + 11) ∴ (6k)2 = –17 k k = 0 or k = − 17 36 The common root α is given by α = 6k − 1 ∴ If k = 0, then α = 0 and if Hence answer is (2) k = − 17 36 then α = 17 . 6 of ‘a’ for which both roots of the equation x 2 − 6ax + 2 − 2a + 9a2 = 0 exceeds 3 are (1) a < 1 (2) a > 1 (3) a < 11 9 (4) a > 11 9 Solution : The equation f (x) ≡ ax 2 + bx + c = 0 has both roots greater than a real no. k if b2 − 4ac > 0 and “ − b > k 2a and a ⋅ f (k ) > 0 ” ∴ a > 1 and “ 6a > 3 2 and 1.(9 − 18a + 2 − 2a + 9a2 ) > 0 ” ∴ a > 1 and 9a2 − 20a + 11 > 0 a > 1 and (9a – 11) (a – 1) > 0 ⎛ a > 11 and a > 1 or a < 11 and a < ⎞ a > 1 and ⎜ 1⎟ ⎝ 9 9 ⎠ ∴ a > 11 . 9 Hence answer is (4) − + 7 (1) 6 ⎛ 7 , 2 ⎞ = 0 , then x is equal to (2) 2 ⎡7 ,2⎤ (3) ⎜ ⎟ ⎝ 6 ⎠ (4) ⎢⎣6 ⎥⎦ Solution : Keep two square roots on one side; since 2x + 3x = 5x, we keep and on one side and on the other side + = On squaring both sides, we get 2x − 3 + 3x − 5 + 2 = 5x − 6 ∴ = 1 6x 2 − 19x + 14 = 0 (6x – 7) (x – 2) = 0 7 x = 2 or 6 Since x = 7 6 does not satisfy the given equation, therefore x = 2 is the only solution. Answer (2) If α, β are the roots of x2 + px + q = 0 and also of x2n + pnxn + qn = 0 and if xn + 1 + (x + 1)n = 0 then n is An integer (2) An odd integer An even integer (4) None of these Solution : If α, β are the roots of x2 + px + q = 0 then α + β = –p and αβ = q. Let xn = t ⇒ when x = α, t = αn and when x = β, t = βn α , β are roots of Therefore, equation x2n + pnxn + qn = 0 reduces to t2 + pnt + qn = 0 with αn + βn = –pn and αnβn = qn α Again, β ⎛ α ⎞n β and α ⎛ α are roots of xn + 1 + (x + 1)n = 0 ⎞n ⇒ ⎜ β ⎟ + 1+ ⎜ β + 1⎟ = 0 ⎝ ⎠ ⎝ ⎠ ⇒ αn + βn + (α + β)n = 0 or –pn + (–p)n = 0 which is true only if n is an even integer. Hence (3) is the correct choice. Example 5 : (1) 5 (2) 4 (3) 3 (4) 2 Solution : Let y = 7 + 7 − 7 + 7 − or y = 7 + 7 − y or y 2 − 7 = 7 − y ...(i) Clearly y 2 − 7 ≥ 0 ⇒ y 2 ≥ 7 ⇒ y ≥ 7 or y ≤ − 7 y ≤ − is not acceptable as y ≥ 0 so only solution may be in the interval y ≥ Since 7 > 2 so y = 2 is not possible Keeping y = 3, 4 and 5 in equation (i) we see That only y = 3 satisfies it and not y = 4 or y = 5. So only solution is y = 3. Answer is (3) then the solution of the equation 16sin2 x + 16cos2 x = 10 is given by x equal to (1) π , π (2) π , π 6 3 3 2 (3) π , π None of these 6 2 Solution : Let 16sin2 x = y , then 16 Hence, y + 16 = 10 y ⇒ y2 – 10y + 16 = 0 or y = 2, 8 cos2 x = 161−sin2 x = 16 y Now 16sin2 x = 2 ⇒ 4 sin2 x = 1 ∴ sin x = ± 1 ⇒ x = π 2 6 Also 16sin2 x = 8 ⇒ 4 sin2 x = 3 ∴ sin x = ± 3 ⇒ x = π 2 3 Hence (1) is the correct answer ❑ ❑ ❑