Quadratic Expressions. The standard form of a quadratic equation is ax + bx + c = 0 where a, b, c R and a 0. The roots of ax + bx + c = 0 are b ± b a 4ac. 3. For the equation ax +bx+c = 0, sum of the roots = - b a, product of the roots =c a. 4. If the roots of a quadratic are known, the equation is x - (sum of the roots)x +(product of the roots)= 0 5. Irrational roots of a quadratic equation with rational coefficients occur in conjugate pairs. If p + q is a root of ax + bx + c =0, then p - q is also a root of the equation. 6. Imaginary or Complex Roots of a quadratic equation with real coefficients occur in conjugate pairs. If p + iq is a root of ax + bx + c = 0. Then p - iq is also a root of the equation. 7. Nature of the roots of ax + bx + c = 0 Nature of the Roots Condition Imaginary b - 4ac < 0 Equal b - 4ac = 0 Real b - 4ac 0 Real and different b - 4ac > 0 Rational b - 4ac is a perfect square a, b, c being rational Equal in magnitude and opposite in sign b = 0 Reciprocal to each other both positive both negative opposite sign c = a b has a sign opposite to that of a and c a, b, c all have same sign a, c are of opposite sign
8.Two equations a x + b x + c = 0, a x + b x + c = 0 have exactly the same roots if a a b c = =. b c 9. The equations a x + b x + c = 0, a x + b x + c = 0 have a common root, if (c a - c a ) = (a b - a b )(b c - b c ) and the common root is ca c a a b a b if a b a b 0. If f(x) = 0 is a quadratic equation, then the equation whose roots are (i) The reciprocals of the roots of f(x) = 0 is f = 0 x (ii) The roots of f(x) = 0, each increased by k is f(x - k) = 0 (iii) The roots of f(x) = 0, each diminished by k is f(x + k) = 0 (iv) The roots of f(x) = 0 with sign changed is f(-x) = 0 (v) The roots of f(x) = 0 each multiplied by k( 0) is f x = 0 k. Sign of the expression ax + bx + c: The sign of the expression ax + bx + c is same as that of a for all values of x if b - 4ac 0 i.e. if the roots of ax + bx + c = 0 are imaginary or equal. If the roots of the equation ax + bx + c = 0 are real and different i.e. b -4ac > 0, the sign of the expression is same as that of a if x does not lie between the two roots of the equation and opposite to that of a if x lies between the roots of the equation.. The expression ax + bx + c is positive for all real values of x if b - 4ac < 0 and a > 0. 3. The expression ax + bx + c has a maximum value when a is negative and x = - b a. Maximum value of the expression = 4 ac b. 4a 4. The expression ax + bx + c, has a minimum when a is positive and x = - b a. Minimum value of the expression = 4 ac b. 4a
Theorems. If the roots of ax + bx + c = 0 are imaginary, then for x R, ax + bx + c and a have the same sign. [NOV-998, APR-999, OCT-993] Proof: The roots are imaginary b 4ac < 0 4ac b > 0 ax + bx + c b c b c b = x + x + = x + + a a a a a 4a b 4ac b = x + + > 0 a 4a For x R, ax + bx + c and a have the same sign. b α = a. If the roots of ax + bx + c = 0 are real and equal to, then α x R, ax + bx + c and a will have same sign. Proof: The roots of ax + bx + c = 0 are real and equal b = 4ac 4ac b = 0 + + = x + x + ax bx c b c a a a b c b = x + + a a 4a b 4ac b = x + + a 4a = x + b a b x = α > 0 for a
For α x R, ax + bx + c and a have the same sign. 3. Let be the real roots of ax + bx + c = 0 and α < β. Then i) x R, α < x < β ax + bx + c and a have the opposite signs ii) x Proof: R, x < α or x > β α, β are the roots of ax + bx + c = 0 > β ax + bx + c and a have the same sign. (APRIL-993, 996) Therefore, ax + bx + c = a( x α)( x β) ax + bx + c = ( x α )( x β ) a i) Suppose x R, α < x < β x α > 0, x β < 0 ax + bx + c < 0 ( x α)( x β ) < 0 a ax + bx + c, a have opposite sign ii) Suppose x, R x < α x < α < β then x α < 0, x β < 0 ( x α)( x β ) > 0 ax + bx + c > 0 a ax + bx + c, a have same sign
Suppose x R, x > β, x > β > α then x α > 0, x β > 0 ( x α)( x β ) > 0 ax + bx + c > 0 a ax + bx + c, a have same sign x R, x < α Or x > β ax + bx + c and a have the same sign. Very Short Answer Questions. Find the roots of the following equations. i) x 7x + = 0 ii) x + x + = 0 iii) 3x + 0x 8 3 = 0 Sol: i)x 7x + = 0 (x 4)(x 3) = 0 x = 4, 3. ii) x + x + = 0 x x = 0 (x )(x + ) = 0 x =,. iii) 3x + 0x 8 3 = 0 0 ± 00 + 3(3) x = 3 0 ± 4 x = 3
0 + 4 0 4 x =, 3 3 x =, 3 3 x =, 4 3 3. Find the nature of the roots of the following equations, without finding the roots. i) x 8x + 3 = 0 ii) 9x 30x + 5 = 0 iii) x x + 3 = 0 iv) x 7x + 0 = 0 Sol: i)x 8x + 3 = 0 Discriminant =b 4ac = 64 4 > 0 Roots are real and distinct. ii) 9x 30x + 5 = 0 Discriminant =b 4ac = 900 4 9 5 = 0 Roots are real and equal. iii) x x + 3 = 0 Discriminant =b 4ac = () 4 3 > 0 Roots are real and distinct. iv) x 7x + 0 = 0 Discriminant =( 7) 4 0 < 0 Roots are imaginary.
3. If α, β are the roots of the equation ax + bx + c = 0, find the value of following expressions in terms of a, b, c. i) + α β ii) + α β iii) α β + α β 4 7 7 4 iv) α β β α, if c 0 v) α + β α + β, if c 0. Sol: i) b c α + β = ; αβ = a a b α + β a b = = αβ c c a ii) α + β ( α + β) αβ = α β α β b = a c a b = c a ca c iii) α β + α β 4 7 7 4 4 4 3 3 = α β ( β + α ) 4 3 = ( αβ) ( ) 3 ( ) α + β αβ α + β 4 3 c b 3c b = 4 + 3 a a a a 4 3 c 3abc b = 4 3 a a
α β iv) β α ( α β ) = α β ( α + β) ( α β) = α β α + β ( ) 4 ( ) = α + β αβ α β b a b 4ac = c a a b (b 4ac) = c a v) c α + β = = α β = + a α β. 4. Find the values of m for which the following equations have equal roots. i) x m(x 8) 5 = 0 ii) (m + )x + (m + 3)x + (m + 8) = 0 iii)(m + )x +(m + 3)x + (m + 5) = 0 Sol: i)x m(x 8) 5 = 0 Discriminant = b 4ac = 0 ( m) 4(8m 5) = 0 4m 3m + 60 = 0 m 8m + 5 = 0 (m 5)(m 3) = 0 m = 3,5
ii) (m + )x + (m + 3)x + (m + 8) = 0 Discriminant = b 4ac = 0 4(m + 3) 4(m + 8)(m + ) = 0 (m + 3) (m + 9m + 8) = 0 6m + 9 9m 8 = 0 3m + = 0 m = 3 iii)(m + )x +(m + 3)x + (m + 5) = 0 Discriminant b 4ac = 0 4(m + 3) 4(m + )(m + 5) = 0 m + 6m + 9 (m + 0m + m + 5) = 0 m 5m + 4 = 0 m + 5m 4 = 0 5 ± 5 + 6 m = 5 ± 4 m = 5. If α and β are the roots of equation x + px + q = 0, form a quadratic equation where roots are (α β) and (α + β). Sol: b c α + β =, αβ = a a [ ] x ( ) ( ) α β + α + β x + ( α β)( α + β ) = 0 x ( ) α + β x + ( α + β) ( ) 4 α + β αβ = 0 b ac b b 4ac x x 0 + = a a a b = p, c = q, a = x (p q)x + p (p 4q) = 0.
6. If x + bx + c = 0, x + cx + b = 0 (b c) have a common root, then show that b + c + = 0. Sol: Given equations are x + bx + c = 0 and x + cx + b = 0 Let α be the common root. Then α + bα + c = 0 α + cα + b = 0 α(b c) + c b = 0 α(b c) = b c α = + b + c = 0 7. Prove that roots of (x a)(x b) = h are always real. Sol: (x a)(x b) = h x (a + b)x + (ab h ) = 0 Discriminant ( ) ( ) ( ) ( ) = (a + b) 4(ab h ) = 0 = a + b 4ab + 4h = a b + 4h = a b + h > 0 Roots are real. 8). Find two consecutive positive even integers, the sum of whose squares in 340. Sol: Let n, n + be the two positive even integers. Then (n) + (n + ) = 340 4n + 4n + 8n + 4 = 340 8n + 8n + 4 = 340 n + n + = 85 n + n 84 = 0 n + n 4 = 0 (n + 7)(n 6) = 0 n = 6
, 4 are two numbers. 9. Find the roots of the equation 4x 4x + 7 = 3x 0x 7. Sol. Given equation can be rewritten as x + 6x + 34 = 0. The roots of the quadratic equation b ± b 4ac ax + bx + c = 0 are. a Here a =, b = 6 and c = 34. Therefore the roots of the given equation are 6 ± (6) 4()(34) 6 ± 00 = () 6 0i (since i ) ± = = = 3+ 5i, 3 5i Hence the roots of the given equation are: 3 + 5i and 3 5i. 0. For what values of m, the equation x ( + 3m)x + 7(3 + m) = 0 will have equal roots? Sol. The given equation has equal roots discriminant is 0. = { ( + 3m) } 4()7(3 + m) = 0 4( + 3m) 8(3 + m) = 0 4(9m 8m 0) = 0 4(m )(9m + 0) = 0 0 36(m ) m + = 0 9 0 m = or m =. 9 Therefore the roots of the given equation are equal iff 0 m, 9.
). If α and β are the roots of ax + bx + c = 0, find the values of α + β and α 3 + β 3 in terms of a, b, c. Sol. From the hypothesis α + β = b c and a αβ = a. α + β = ( α + β) ( αβ) b c = a a b ac b ac = = a a a And 3 3 α + β = ( α + β)( α + β αβ ) = ( α + β)[( α + β) αβ αβ] = ( α + β) ( ) 3( ) α + β αβ b b c = 3 a a a 3 b b 3c 3abc b = = 3. a a a a. Form a quadratic equation whose roots are 3 5 and 3 5. Sol. Let α = 3 5 and β = 3 5. Then α + β = ( 3 5) + ( 3 5) = 0 and αβ = ( 3 5)( 3 5) = ( 3)( 3) ( 3)(5) 5( 3) 5( 5) = 0 3 + 0 3 + 5 = 3. The required equation is x ( α + β )x + αβ = 0 x ( 0)x + 3 = 0. i.e., x + 0x + 3 = 0
3. Let α and β be the roots of the quadratic equation ax + bx + c = 0. If c 0, then form the quadratic equation whose roots are α and β. α β Sol. From the hypothesis we have α + β = b c and a αβ = a. Since, c 0, α 0 and β 0 α β β( α ) + α( β) + = α β αβ α + β αβ = αβ b α + β a = = αβ c a b b = = + a c And α β ( α + β ) + αβ = α β αβ b c + a a a + b + c = =. c c a Therefore b a + b + c x + x + = 0 c c is the required equation. 4. Find a quadratic equation, the sum of whose roots is and the sum of the squares of the roots is 3. Sol. Let α and β be the roots of a required equation. Then α + β = and α + β = 3. Since ( ) ( ), αβ = α + β α + β () 3 6. αβ = =
Therefore x ( α + β )x + αβ = 0 becomes This is a required equation. x x 6 = 0. 5. Find all numbers which exceed their square root by. Sol. Let x be the number satisfying the given condition. Then x = x + x = x... () On squaring both sides and simplifying we obtain x 4x + 44 = x x 5x + 44 = 0 x(x 6) 9(x 6) = 0 (x 6)(x 9) = 0 (x ) = x The roots of the equation (x 6)(x 9) = 0 are 9 and 6. But x = 9 does not satisfy equation (), while x = 6 satisfies (). Therefore the required number is 6. 6. Prove that there is a unique pair of consecutive positive odd integers such that the sum of their squares is 90 and find it. Sol. Let x and x+ be the two consecutive positive odd integers. x + (x + ) = 90...() x + x + 4x + 4 = 90 x + 4x 86 = 0 x + x 43 = 0 x + 3x x 43 = 0 x(x + 3) (x + 3) = 0 (x + 3)(x ) = 0 x { 3,} Hence is the only positive odd integer satisfying equation (). Therefore (, 3) is the unique pair of integers which satisfies the given condition.
7. Find the quadratic equation whose roots are, 5. Sol: The quadratic equation whore roots are,5 is x -(sum of the roots) x+ product of the roots =0 x ( + 5)x +.5 = 0 i.e., x 7x + 0 = 0. ii. Find the quadratic equation whose roots are 3 +, 3 +. Sol: The quadratic equation is x -(sum of the roots) x+ product of the roots =0 x (3+ + 3 ) x + (3 + )(3 ) = 0 i.e., x 6x + 7 = 0. iii. Find the quadratic equation whose roots are + 3i, 3i. Sol: The quadratic equation is x -(sum of the roots) x+ product of the roots =0 x ( + 3i + 3i)x + ( + 3i) ( 3i) = 0 i.e., x 4x + 3 = 0. iv. Find the quadratic equation whose roots are, 7. 3 Sol: 7 7 x x + ( ) = 0 3 3 i.e., 3x + 0x + 7 = 0 v. Find the quadratic equation whose roots are ± 3. Sol: 3 3 3 3 x + x + = 0 9 x = 0 x 9 = 0
vi. Find the quadratic equation whose roots are 3 ± i 5. Sol: 3 + i 5 3 i 5 3 + i 5 3 i 5 x + x 0 + = x 7 3x + = 0 x 6x + 7 = 0 vii. Find the quadratic equation whose roots are r q, q p. Sol: x r q r q x + = 0 q p q p pqx + ( q rp) x rq = 0 8. If the quadratic equations ax + bx + c = 0 and ax + cx + b = 0, (b c) have a common root then show that a + 4b+4c = 0. Sol: ax + bx + c = 0 ax + cx + b = 0 Let α be a common root. aα + bα + c = 0 aα + cα + b = 0 α(b c) + c b = 0 (α )(b c) = 0 α = a b + + c 4 a + 4b + 4c = 0
9. If x 6x + 5 = 0 and x x + p = 0 have a common root, then find p. Sol: x 6x + 5 = 0 (x 5)(x ) = 0 x =, 5 Now x x + p = 0 can have or 5 as root then + p = 0 p = (or) 5 60 + p = 0 p = 35. 0. If x 6x + 5 = 0 and x 3ax + 35 = 0 have a common root, then find a. Sol: x 6x + 5 = 0 (x 5)(x ) = 0 x = 5, Now, 5, satisfy x 3ax + 35 = 0 5 5a + 35 = 0 60 = 5 a a = 4 3a + 35 = 0 3a = 36 a =.. If the equations x + ax + b = 0 and x + cx + d = 0 have a common root and the first equation have equal roots, then prove that (b + d) = ac. Sol: x + ax + b = 0 x + cx + d = 0 x + ax + b = 0 have equal roots. a 4b = 0
a = 4b x + ax + b = 0 x + cx + d = 0 α(a c) + b d = 0 as α = a a α = a a + c + d = 0 4 a ac + 4d = 0 4b + 4d = ac or (b + d) = ac.. Discuss the signs of the following quadratic expressions when x is real. i) x 5x + 4 ii) x x + 3 Sol: i)f(x) = x 5x + 4 f (x) = (x 4)(x ) 5 5 y = x + 4 4 5 9 y = x 4 9 5 y + = x 4 4 (5/, 9/4) f(x) < 0 < x < 4
f(x) > 0 x > 4 and x < ( < x < ) (4 < x < ) ii) x x + 3 f(x) = x x + 3 f (x) = x + 3 4 y = x + 4 y = x 4 (/,/4) f(x) > 0 x Real numbers. 3. For what values of x the following expressions are positive? i) x 5x + 6 ii) 3x + 4x + 4 iii)4x 5x + Sol: i)f(x) = x 5x + 6 f(x) > 0 x 5x + 6 > 0 (x 3)(x ) > 0 x > 3 or x < ( < x < ) (3 < x < )
ii) 3x + 4x + 4 f(x) = 3x + 4x + 4 f(x) > 0 3x + 4x + 4 > 0 4 4 x + x + > 0 3 3 4 4 x + + > 0 3 3 9 8 x + + > 0 x R. 3 9 iii)f(x) = 4x 5x + = 5x 4x 4 = 5 x x 5 5 4 = 5 x 5 5 5 4 = 5 x 5 5 4x 5x + > 0 5x 4x < 0 4 ± 6 + 40 x = 0 4 ± 4 x = 0 ± 4 x = 5 4 + 4 < x < 5 5
4. For what values of x, the following expressions are negative? i) x 7x + 0 ii) 5 + 4x 3x iii)x + 5x 3 Sol: i)f(x) = x 7x + 0 f(x) < 0 (x 5)(x ) < 0 < x < 5 ii) f(x) = 5 + 4x 3x f(x) < 0 5 + 4x 3x < 0 3x 4x 5 > 0 3x 9x + 5x 5 > 0 3x(x 3) + 5(x 3) > 0 (x 3)(3x + 5) > 0 5 x > 0, x < 3 5 < x < (3 < x < ) 3 iii)f(x) = x + 5x 3 f(x) < 0 x + 5x 3 < 0 x + 6x x 3 < 0 x(x + 3) (x + 3) < 0 (x )(x + 3) < 0 3 < x <.
5. Find the changes in the sign of the following expressions and find their extreme values. i) x 5x + 6 ii) 5 + 4x 3x Sol: i)f(x) = x 5x + 6 f(x) = (x 3)(x ) f(x) < 0 < x < 3 And f(x) > 0 ( < x < ) (3 < x < ) 5 6 5 y = x + 4 y min = 4 ii) f(x) = 5 + 4x 3x f(x) < 0 5 + 4x 3x < 0 3x 4x 5 < 0 3x 9x + 5x 5 < 0 3x(x 3) + 5(x 3) < 0 (x 3)(3x + 5) < 0 3 < x < 3 5 f(x) > 0 3 < x < (3 < x < ) 5 y = 5 + 4x 3x y = 3 x x 5 3 4
4 y = 3 x 5 3 9 49 y = 3 x + 3 3 9 49 y = 3 x + 3 3 y min 49 = 3 6. Find the maximum or minimum of the following expression as x varies over R. i) x x + 7 iii)x + 5 3x iv) ax + bx + a (a, b R and a 0) Sol: i)y = x x + 7 y = x + 7 4 7 y = x + 4 y min max 7 = 4 ii) f(x) = x + 5 3x 5 y = 3 x x 3 3 5 y = 3 x 3 9 9 6 y = 3 x + 3 9 y 6 = 3
iii)f(x) = ax + bx + a y = a x + x + a b b b y = a x + + a 4a b 4a b y = a x + + a 4a b b 4a y = a x + a 4a min b + 4a y max =,a < 0 4a y when a > 0. 7.. Solve the following in equations by algebraic method. i) 5x + 4x 4 0 ii) x x + < 0 iii) 3x x 0 Sol: i)5x + 4x 4 0 5x + 0x 6x 4 0 (3x + )(5x ) 0 x + x 0 3 5 x 3 5 ii) x x + < 0 (x ) < 0 Not possible. (x ) 0 No solution.
iii) 3x x 0 x + 3x 0 x + 4x x 0 x(x + ) (x + ) 0 (x )(x + ) 0 x. Short Answer Questions. If x, x are the roots of equation ax + bx + c = 0 and c 0. Find the value of (ax + b) + (ax + b) in terms of a, b, c. Sol: ax + bx + c = 0 b c x + x = & x. x = a a ax + bx + c = 0 ( ) x ax + b = c ax c + b = x c Similarly ax + b = x x + x (ax + b) + (ax + b) = c = (x + x ) x x c b ac = c a b ac = a c
. If α, β are the roots of equation ax + bx + c = 0, find a quadratic equation whose roots are α + β and α + β. Sol: x x α + β + + + α β ( ) α + β + = 0 α β ( α + β) αβ ( α + β) αβ x x ( α + β) αβ + + = 0 ( αβ) ( αβ) b ac (b ac)a b ac x x + + a = 0 a c c a c x (b ac)(a + c )x + (b ac) = 0 3. Solve the following equation 4 3 x + x x + x + = 0. Sol: 4 3 x + x x + x + = 0 4 3 x + x 0x x + x + = 0 x (x + x 0) (x x ) = 0 [ ] [ ] x (x + 5)(x ) (x )(x + ) = 0 (x ) x (x 5) x + = 0 3 (x )[x + 5x x ] = 0 (x )[(x )(x + 3x + )] = 0 x =, ;x + 3x + = 0 3 ± 9 4 x = 3 ± 5 x = 4. Sol: + x x 3 + 3 = 0 Let 3 x = t 3 3 t + = 0 t 3t + 3 0t = 0 3t 9t t + 3 = 0
3t(t 3) (t 3) = 0 (3t )(t 3) = 0 t =, t = 3 3 x x 3 = 3,3 = 3 x =, 5. x x 3 5 + = x 3 x When x 0, x 3. Sol: x x 3 = t 5 t + = t t 5t + = 0 t 4t t + = 0 t(t ) (t ) = 0 (t )(t ) = 0 t = ;t = x = 4 x 3 x = 4x 3x = x = 4 (Or) x = x 3 4 4x = x 3 3x = 3 x = 6. Sol: x + 7 x + + 5 = 0, when x 0. x x x + = t x t 7t + 5 = 0 t 5t t + 5 = 0 t(t ) 5(t ) = 0
5 t =, t = but x + x R x 5 x + = only possible x x 5x + = 0 (x )(x ) = 0 x =, + Now if x + = x x x + = 0 ± 4 x = ± 3i x = 7. Sol: x + 5 x 6 0 + + =, when x 0. x x x + 5 x 6 0 + + = x x x + = t x t 5t + 4 = 0 (t 4)(t ) = 0 t = 4, x + = 4 x x 4x + = 0 4 ± 6 4 x = x = ± 3 (Or) x + = x x x + = 0 ± 4 x = ± 3i x =
8. Find a quadratic equation for which the sum of the roots is 7 and the sum of the squares of Sol: the roots is 5. α + β = 7, α + β = 5 ( α + β) αβ = 5 49 αβ = 5 4 = αβ αβ = x 7x + = 0. 9. Determine the range of the following expressions. i) x + x + x x + x + ii) x + 3x + 6 (x )(x + ) iii) x + 3 x 6x + 5 iv) x 3x + Sol: i) y = x + x + x x + yx yx + y = x + x + x (y ) + x( y ) + y = 0 Discriminant 0 ( y ) 4(y )(y ) 0 (y + y + ) 4(y y + ) 0 3y + 0y 3 0 3y 0y + 3 0 3y 9y y + 3 0
3y(y 3) (y 3) 0 (3y )(y 3) 0 y 3 3 x + ii) y = x + 3x + 6 x y + 3xy + 6y x = 0 x (y) + x(3y ) + 6y = 0 Discriminant 0 (3y ) 4(y)(6y ) 0 9y 6y + 48y + 6y 0 39y + 0y + 0 39y 0y 0 39y 3y + 3y 0 3y(3y ) + (3y ) 0 (3y + )(3y ) 0 y. 3 3 iii) y = (x )(x + ) x + 3 x + x y = x + 3 yx + 3y = x + x x + x( y) 3y = 0 Discriminant 0 ( y) + 4( + 3y) 0 y y + + 8 + y 0 y + 0y + 9 0
(y + )(y + 9) 0 y or y 9 ( < y 9) ( y < ) iv) y = x 6x + 5 x 3x + y(x 3x + ) = x 6x + 5 x (y ) + x( 3y + 6) + y 5 = 0 Discriminant 0 (6 3y) 4(y )(y 5) 0 9y 36y + 36 8y + 36y 40 0 y 4 0 (y )(y + ) 0 y or y ( < y ) ( y < ) 0. Prove that + 3x + x + (3x + )(x + ) does not lie between and 4 if x R. Sol: x + + 3x + y = (3x + )(x + ) 4x + y = 3x + 4x + 3yx + x(4y 4) + y = 0 Discriminant 0 4(y ) 4 3y(y ) 0 6(y ) y(y ) 0 4(y )[4(y ) 3y)] 0
4(y )(y 4) 0 (y )(y 4) 0 y 4 or y. x. If x is real, prove that x 5x + 9 lies between and. x Sol: y = x 5x + 9 y(x 5x + 9) = x x y + x( 5y ) + 9y = 0 Discriminant 0 ( 5y ) 4 9y 0 5y + 0y + 36y 0 y + 0y + 0 y y + y 0 y(y ) + (y ) 0 (y )(y + ) 0 y x p. In the expression x 3x + takes all values of x R, then find the bounds for p. x p Sol: y = x 3x + y(x 3x + ) = x p x y + x( 3y ) + y + p = 0 Discriminant 0 ( 3y ) 4y(y + p) 0 9y + 6y + 8y 4p 0 y + y(6 4p) + 0 Discriminant < 0
(6 4p) 4 < 0 6p 48p + 36 4 < 0 6p 48p + 3 < 0 p 48p + 3 < 0 p 3p + < 0 (p )(p ) < 0 < p <. 3. If c ab and the roots of (c ab)x (a bc)x + (b ac) = 0 are equal, then show that a 3 + b 3 + c 3 = 3abc or a = 0. Sol: Roots are equal Discriminant = 0 4(a bc) 4(c ab)(b ac) = 0 4 3 3 a + b c a bc b c + c a + b a a bc = 0 4 3 3 a a bc + c a + b a a bc = 0 3 3 3 a[a + b + c 3abc] = 0 3 3 3 a + b + c 3abc = 0 or a = 0. 4. Solve the following in equations by graphical method. i) x 7x + 6 > 0 ii)5x + 4x 4 0. Sol: i) x 7x + 6 > 0 (x 6)(x ) > 0 6
ii)5x + 4x 4 0 5x + 0 x 6x 4 0 5x(3x + ) (3x + ) 0 x 3 5 /3 /5 (0, 4) 5. Solve the following in equations. i) 3x 8 < ii) x + 6x 5 > 8 x. Sol: i) 3x 8 < Possible when 3x 8 > 0 8 x > 3 Also 3x 8 0 Solution does not exist. ii) x + 6x 5 > 8 x Possible x + 6x 5 0 x 6x + 5 0 (x 5)(x ) 0 x 5 ()
Squaring on both sides we get x + 6x 5 > 64 + 4x 3x 0 > 5x 38x + 64 + 5 or 5x 38x + 69 < 0 5x 3x 5x + 69 < 0 5x(x 3) 3(x 3) < 0 (x 3)(5x 3) < 0 3 3 < x < () 5 Intersection of () and () 3 < x 5. 6. The cost of a piece of cable wire is Rs.35/-. If the length of the piece of wire is 4 meters more and each meter costs Rs./- less, the cost would remain unchanged. What is the length of the wire? Sol. Let the length of the piece of wire be l meters and the cost of each meter br Rs.x/-. By the given conditions lx = 35 Also, (l + 4)(x ) = 35...() i.e., lx l + 4x 4 = 35...() From () and (), 35 l + 4x 4 = 35 i.e., 4x = l + 4 Therefore + 4 x = l. 4 On substituting this value of x in () and simplifying, we get l +4 l = 35 4 l + 4l 40 = 0 l + 4l 0l 40 = 0 l( l + 4) 0( l + 4) = 0 ( l + 4)( l 0) = 0
The roots of the equation ( l + 4)( l 0) = 0 are 4 and 0. Since the length cannot be negative, l = 0. Therefore the length of the piece of wire is 0 meters. 7. Suppose that the quadratic equations ax + bx + c = 0 and bx + cx + a = 0 have a common root. Then show that a 3 + b 3 + c 3 = 3abc. Sol. The condition for two quadratic equations common root is = (c a c a ) (a b a b )(b c b c ) a x + b x + c = 0 and a x + b x + c = 0 to have a Here a = a,b = b,c = c,a = b, b = c and c = a. Therefore (cb a ) = (ac b )(ba c ) 4 i.e., b c a bc + a = 3 3 a bc ac b a + b c 4 3 3 i.e., a + ab + ac = 3a bc Hence 3 3 3 a + b + c = 3abc (since a 0). 8. Find the maximum or minimum value of the quadratic expression (i) x 7 5x (ii) 3x + x + Sol. i) Comparing the given expression with ax + bx + c, We have a = 5, b = and c = 7. So, 4ac b 4( 5)( 7) () 34 = = 4a 4( 5) 5 And b = =. a ( 5) 5 Since a < 0, x 7 5x has absolute maximum at x = 5 and the maximum value is 34. 3 ii) Here a = 3, b = and c =. So 4ac b 4(3)() () 3 = = and 4a 4(3) 3 b = =. a 6 3
Since a > 0, 3x + x + has absolute minimum at x = and the minimum value is 3 3 3. 9. Find the changes in the sign of 4x 5x + for x R and find the extreme value. Sol. Comparing the given expression with ax + bx + c, we have a = 5 < 0. The roots of the equation 5x 4x = 0 are ± 4 5. Therefore, when 4 + 4 < x <, the sign of 4x 5x + is positive and when 5 5 4 4 x < or x > +, the sign of 4x 5x + is negative. 5 5 Since a < 0, the maximum value of the expression 4x 5x + is 4ac b 4( 5)() (4) = 4a 4( 5) 56 4 = =. 0 5 Hence the extreme value of the expression 4x 5x + is 4 5. 0. Show that none of the values of the function Sol. Let y 0 be a value of the given function. Then x0 R such that x + 34x 7 x + x 7 over R lies between 5 and 9. y 0 0 0 = 0 0 x + 34x 7. x + x 7 If y 0 =, then clearly y 0 (5,9). Suppose that y0. Then the equation x 0 is a real root of it. Therefore 0 0 0 y (x + x 7) = x + 34x 7 is a quadratic equation and (y )x + (y 34)x (7y 7) = 0 is a quadratic equation having a real root x 0. 0 Since all the coefficients of this quadratic equation are real, the other root of the equation is also real.
Therefore 0 0 0 = (y 34) + 4(y )(7y 7) 0. On simplifying this we get 0 0 y 4y + 45 0 i.e., (y0 5)(y0 9) 0 Therefore y0 5 or y0 9. Hence y 0 does not lie in (5, 9). Hence none of the values of the given function over R lies between 5 and 9.. Find the set of values of x for which the inequalities x 3x 0 < 0, 0x x 6 > 0 hold simultaneously. Sol. We have x 3x 0 = (x + )(x 5). Hence and 5 are the roots of the equation x 3x 0 = 0. Since the coefficient of x in the quadratic expression x 3x 0 is positive. x 3x 0 < 0 < x < 5. We have 0x x 6 = (x )(x 8). Hence and 8 are the roots of the equation 0x x 6 = 0. Since the coefficient of x in the quadratic expression 0x x 6 is negative. 0x x 6 > 0 < x < 8. Hence x 3x 0 < 0 and 0x x 6 > 0 x (,5) (,8) Therefore the solution set is {x R : < x < 5}.
. Solve the in equation x + > 8 x. Sol. When a, b R and a 0,b 0 then a > b a > b 0. Therefore x + > 8 x (x + ) > (8 x ) 0 and x > x <. 3 0 3 We have Now (x + ) > (8 x ) x + x 6 > 0. x + x 6 = (x + 3)(x ) Since 3 and are the roots of the equation x + x 6 = 0 and the coefficient of x in x + x 6 is positive, x + x 6 > 0 x (, 3) (, ) We have 8 x 0 x 8 x x, Also x + > 0 x > Hence x + > 8 x 0 ( ) x (, 3) (, ), and x > x (, ) Therefore the solution set is { x : < x } R.
3. Solve the in equation (x 3)( x) < 4x +x +. Sol. The given in equation is equivalent to the following two inequalities. (x 3)( x) 0 and (x 3)( x) < 4x + x +. (x 3)( x) 0 (x )(x 3) 0 x 3 x + 5x 6 < 4x + x + 5x + 7x + 7 = 0 The discriminate of the quadratic expression 5x + 7x + 7 is negative. Hence 5x + 7x + 7 > 0 x R. Hence the solution set of the given in equation is {x R : x 3}.