EQUATIONS REDUCIBLE TO QUADRATICS EQUATIONS
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1 CHAPTER EQUATIONS REDUCIBLE TO QUADRATICS EQUATIONS Numbers The numbers,, are called natural numbers or positive integers. a is called a fraction where a and b are any two positive integers. b The number system consisting of positive and negative fractions is called rational number. The rational number is of the form b a where a and b are integers and b 0. That is 4,,, 0,,,, 4 are called integers. The number,, 5 which cannot be epressed as b a (i.e. ratio of two integers), are called irrational numbers. The rational numbers, positive and negative, irrational numbers and zero constitute the real number. Types of Numbers Natural number or N,,, 4,..
2 Remedial Mathematics for B. Pharmacy Real number or R The set of rational and irrational numbers i.e., R = Q I r, where I r denotes the Irrational numbers. Rational number or Q.. 4,, 0,,.. Primes or P,, 5, 7,,.. Composite numbers 4, 6, 8, 9, 0,.. Even numbers, 4, 6, 8, 0,.. Odd numbers,, 5, 7, 9,.. Comple Numbers The numbers of the form + iy where i = and a and b are real numbers are called comple numbers. Useful Formulae (a + b) = a + b + ab (a b) = a + b ab (a + b) + (a b) = a + b (a + b) (a b) = 4ab (a + b) (a b) = a b (a + b) = a + b + a b + ab (a + b) = a + b + ab (a + b) (a b) = a b a b + ab (a b) = a b ab (a b) (a + b ) = (a + b) ab (a + b) a + b = (a + b) (a ab + b ) a b = (a b) + ab (a b) a b = (a b) (a + ab + b ) ( + a) ( + b) = + (a + b) + ab ( a) ( b) = (a + b) + ab
3 Equations Reducible to Quadratics Equations Quadratics Equation Definition The equation of the form a + b + c = 0 is called a quadratic equation, where a, b, c are real numbers and a 0. Eample: (i) = 0 (ii) + = 0 Quadratic Equation of a Roots It has only two roots, the value of the given equation is called a root. Such as = = 0 using factorization 5 ( ) ( ) = 0 (5 ) ( ) = 0 = 5 and = Hence roots are, 5 Working Rule Given quadratic equation a + b + c = 0 can be solved as b b 4ac = a b b 4ac b b 4ac Let =, = a a Eample : Solve the quadratic equation = 0 Let f() = = 0 using factionzation = 0 ( 5) 4 ( 5) = 0 ( 5) ( 4) = 0 Hence solutions, = 5, 4
4 4 Remedial Mathematics for B. Pharmacy Eample : Solve the quadratic equation 5 = 0 we know that a b = (a b) (a + b) According to problem 5 = 0 () (5) = 0 ( 5) ( + 5) = 0 Hence roots are = 5, 5 Eample : Find the value of given quadratic equation is 5 + = 0 Given equation 5 + = 0 by quadratic equation a + b + c = 0 Now, a =, b = 5, c =, put values = b b 4ac a = ( 5) ( 5) () 4()() = = 5 7 Hence = 5 7 and = 5 7 Eample 4: Solve the quadratic equation = 0 we know that formula = b b 4ac a given values a =, b = 4, c = 5 = () ( 5) () = = 6 56
5 = = = = = 6 Hence roots are, 5 Equations Reducible to Quadratics Equations 5 0 = = 5 6 Problem = 0. = 0 Ans:,. 6 = = 4 4 Ans: 7, Ans: 4, 4 Ans: ( +, ) Sum of the Roots: If a + b + c = 0 then sum of the roots + = a b Product of the Roots: If a + b + c = 0 then product of the roots () = a c where are two roots and a, b, c are real numbers Eample : Find the value of and find the sum and product of the roots of given equation = 0 According to problem given equation is = 0
6 6 Remedial Mathematics for B. Pharmacy using formula b b 4ac = a a = 5, b = 5, c = put values in equation () 5 5 = (5) = (5) Now sum of roots, + = a b = 4 (5) () () = 5 = 5 c Product of the roots = a = 5 Eample : Find the sum of the roots and product of the roots = 0 a =, b = 5, c = 0 Sum of the roots, + = a b 5 = = 5 c product of the root (= a = 0 = 0 Hence required sum of the roots 5 and product of the roots 0 Eample : Find the set of values of for given equation has real roots. 6 = 0 We know that quadratic equation a + b + c = 0
7 Equations Reducible to Quadratics Equations 7 b b 4ac = a a =, b = 6 and c =, put values in equation () ( 6) ( 6) 4 ( ) ( ) = ( ) = the given equation will have real roots if > 0. = b 4ac = i.e > 0 i.e. 8 > 6 9 i.e. >..() 9 the required set of values = R : Where R is the set of real numbers. Similar proof or solution method same. (i) = 0 (ii) = 0 Nature of roots Some important cases of finding roots of the quadratic equations. a + b + c = 0. The roots and are real and distinct, if > 0 i.e. if b 4ac > 0.. The roots and are real and equal ( = ), if = 0 i.e. if b 4ac = 0. The roots and are imaginary and distinct, if < 0 i.e. b 4ac < 0 4. The roots are rational and distinct, if b 4ac is a perfect square Eample 4: If, be the roots of the equation 5 + = 0, find the value of β α + Given equation 5 + = 0
8 8 Remedial Mathematics for B. Pharmacy factorization 4 + = 0 ( ) ( ) = 0 ( ) ( ) = 0 = 0 and = 0 = and = or = and = According to problem α + = β 4 = = 6 7 = 4 4 Eample 5: solve = 5 the given equation can be written as + 5 = squaring the both sides ( + 5) = = = = = 0 ( + 8) + ( + 8) = 0 ( + ) ( + 8) = 0 + = 0 and + 8 = 0 = and = 8 Hence roots are {, 8}
9 Formation of Quadratic Equation with Given Roots Equations Reducible to Quadratics Equations 9 If, be the given roots, then the required quadratic equation is ( + ) + = 0 or (sum of the roots) + product of the roots = 0 Eample 6: Find the equation whose roots are + and Let = + and = Solve the following Now, if, are any root of comple number then equation will be ( + ) + = 0 [ + + ] + ( + ) ( ) = () ( ) = 0 a b = (a b) (a + b) = = 0 Problem.. = 4. ( ) ( + 7) = 0. 5 = = = Ans: (0, 4) Ans: (, 7) Ans: (0, 5) Ans: ( 4, ) = 0 Ans: Ans:, 5 5,
10 0 Remedial Mathematics for B. Pharmacy 7. ( + ) = 5 Ans:, = = 5 5. = = 0. ( + 5) = 5 4. (6 ) = 5 5. ( + ) = 6 7 =, 0, 4 Ans: (, 6) Ans: 4, Ans: (5, 8) Ans:, Ans:, Ans: 5, Ans: Ans:, , Equations Reducible to Quadratic Equations of Different Types Type-I a n + b n + c = 0 Method: In this type of equations, we put n = y, so that y = ( n ) = n and the given equation becomes ay + by +c = 0, which is a quadratic equation
11 Equations Reducible to Quadratics Equations Eample : Solve 7 The given equation can be written as 7 0 Putting = y in () y 7y + = 0 y y 4y + = 0 y (y ) 4 (y ) = 0 (y ) (y 4) = 0 y =, or y = 4, return value y = = = = 7 and y = 4 = 4 = 4 = 64 solution set is {7, 64} Eample : Solve + = 0 putting = y in () + = 0 y + y = 0 y + y y = 0 y (y + ) (y + ) = 0 (y ) (y + ) = 0 y = 0 or y + = 0 y = or y =, return value then y = = = = y = = = = 8 Hence solution set is {, 8}...()..()
12 Remedial Mathematics for B. Pharmacy Eample : Solve = 0..() Putting = y in () y 9y + 8 = 0 y 8y y + 8 = 0 y (y 8) (y 8) = 0 (y ) (y 8) = 0 y = 0 or y 8 = 0 y = or y = 8, return value then y = = = y = 8 = 8 = = Hence solution is {, }. Eample 4: Solve for 8 0,..() Putting = y in () y y 8 = 0 y 6y + y 8 = 0 y (y 6) + (y 6) = 0 (y + ) = 0 or y 6 = 0 y = or y = 6, return the values y = = + 4 = = 4 y = 6 = 6 = = 6 = 5 Hence solution is,
13 Equations Reducible to Quadratics Equations Eample 5: Solve for, 5 Putting 4 = 4 = y and squaring the both sides, we get 4 = y The given equation becomes y 5y + 4 = 0 y 4y y + 4 = 0 y (y 4) (y 4) = 0 (y ) (y 4) = 0 y = 0 or y 4 = 0 y = or y = 4, return the values when y = = = = For (+) For ( ) = + = = 4 5 = and = 5 y = 4 = 4 = =
14 4 Remedial Mathematics for B. Pharmacy or = ( + ) For (+) For ( ) = = 4 6 = 7 7 = 5 = 7 5 = 7 5 Hence 4,, 7,. 5 7 Eample 6: Solve 6 + = 7 The given equation can be written as 6 7 Putting = y in equation () = 0..() 6y 7y + = 0 6y y 4y + = 0 y (y ) (y ) = 0 (y ) (y ) = 0 y = 0 or y = 0 y = or y =, return the values y = =, squaring the both sides = 4 y = =, squaring the both sides = 9 4 Hence the solution is,
15 Equations Reducible to Quadratics Equations 5 b Type-II a + = c, 0 where a, b, c are constants. Eample : Solve for + =, 0, 6 Putting + = 6 = y, the given equation becomes y + = y 6 6y + 6 = y 6y y + 6 = 0 6y 9y 4y + 6 = 0 y (y ) (y ) = 0 (y ) (y ) = 0 y = 0 or y = 0 y = or y =, return the values when y = then = = 4 4 = 4 4 = =, squaring the both sides and y = then 9 = 4 =, squaring the both sides
16 6 Remedial Mathematics for B. Pharmacy 4 = 9 9 = 9 = 9 Hence, the solution set are 9, 4 5 Eample : Solve 5 = 8, 0 The given equation can be written as 5 5 = = = 0 5 ( 5) + 7 ( 5) = 0 (5 + 7) ( 5) = = 0, 5 = 0 7 = or = Hence the solution set are, 5. 5 Eample : Solve + = 0 The given equation can be written as + = = = 0 ( ) ( ) = 0 ( ) ( ) = 0 = 0 and = 0 = and = = Hence the solution set are,.
17 Equations Reducible to Quadratics Equations 7 Type-III Reciprocal Equations P q + r = 0 Where p, q, r are constants. Put = y then the given equation becomes p (y ) qy + r = 0. Eample : Solve = 0, 0 Put = y = y = y + + = y + = y Putting these value in given equation (y ) 6 (y) + 6 = 0 y 6 6y + 6 = 0 y 6y + 0 = 0 y 0y 6y + 0 = 0 y (y 0) (y 0) = 0 (y 0) (y ) = 0 y 0 = 0 and y = 0 y = 0 0 y =, y =
18 8 Remedial Mathematics for B. Pharmacy 0 when y = when y = 0 then + = then = 0 = = by cross multiplication + = = 0 ( ) = = 0 = 0, ( ) = 0 ( ) ( ) = 0 =, ( ) ( ) = 0 = 0, = 0 = and = Hence =,,,. Eample : Solve = 0, 0 the given equation can be written as = 0 Put = y, then given equation becomes (y ) 9y + 4 = 0 y 9y + 0 = 0 y 5y 4y + 0 = 0 y (y 5) (y 5) = 0 (y 5) (y ) = 0 y 5 = 0 or y = 0 y = 5, y =
19 Equations Reducible to Quadratics Equations 9 when y = 5 when y = then 5 = then = 5 = = + = 5 + = = 0 + = = 0 ( ) ( ) = 0 ( ) ( ) = 0 ( ) ( ) = 0 ( ) ( ) = 0 =, =, = Hence =,,, Problem. Solve = = = 0 Ans:,, 5, 5 Ans:,, Ans: 5,,,, 4 4 5
20 0 Remedial Mathematics for B. Pharmacy Type-IV Equation of the Form Pm + qm + r = 0 Put m = y, then the given equation becomes py + qy + r = 0 or a + b + c = 0 which is quadratic Type-V Equation of the Form ( + a) ( + b) ( + c) ( + d) + K = 0 where a, b, c, d, k are real constants. Type-VI Irrational equations reducible to quadratics as ( a b) = k or ( a b) + ( c d) = k or ( a b) + ( c d) = ( e f ) Type-VII Problem : Simultaneous equations in two variables Solve for = 0 The given equation can be written as ( ) = = = = 0 Let = y and squaring the both sides = y 8y 8y + = 0 9y (9y ) (9y ) = 0 (9y ) (9y ) = 0 9y = 0 or 9y = 0 y = or y = 9 9 = 9 or = 9 = or = Base same then power equal = or = Hence =
21 Equations Reducible to Quadratics Equations Problem : Solve + + = 0 The given equation can be written as. + = 0 by eponent rule..() Let = y, put in equation () y. + y = 0 or 9y + = 0 y 9y 0y + = 0 9y 9y y + = 0 9y (y ) (y ) = 0 (9y ) (y ) = 0 9y = 0 and y = 0 y = and y = 9 = and = 9 = = o = and = 0 Hence = 0, Problem : Solve for ( + ) ( + ) ( + ) ( + 4) + = 0 Type - V [( + ) ( + )] [( + ) ( + 4)] + = 0 [ ] [ ] + = 0 ( + + ) ( ) + = 0 Which is not similar, so the given equation can be written as [( + ) ( + 4)] [( + ) ( + )] + = 0 ( ) ( ) + = 0..() Let + 5 = y, put in () (y +4) (y + 6) + = 0 y + 4y + 6y = 0 y + 0y + 5 = 0
22 Remedial Mathematics for B. Pharmacy y + 5y + 5y + 5 = 0 y (y + 5) + 5 (y + 5) = 0 (y + 5) (y + 5) = 0 (y + 5) = 0 y = = y Now + 5 = = 0 = () (5) () = 5 5 Problem 4: Solve = 8..() Suppose 4 6 = y so that 4 6 = y and so 4 = y + 6, putting these values in () y + y + 6 = 8 y + y = 0 y + 4y y = 0 y (y + 4) (y + 4) = 0 (y ) (y + 4) = 0 y = 0 and y + 4 = 0 y = and y = = and 4 6 = 4 squaring the both sides 4 6 = 9 and 4 6 = = 0 and 4 = = 0 and = 4 ( 4) () 4 () ( ) ( ) + 5( ) = 0 and =
23 Equations Reducible to Quadratics Equations The solution set is 4 80 ( ) ( + 5) = 0 and = = and = = 70 = ,,,. Problem 5: Solve ( 5 ) = Type-VI We have ( 5 ) = +, squaring both the sides = = 0 = 0 ( ) = 0 = 0, = 0 = 0 or = Hence the solution set 0, Problem 6: = 5 The given equation can be written as 9 = 5 4, squaring the both sides 9 + = (5) 4 0 = 0 4 ( 0) = = 5 4, squaring the both sides = 5 (4 )
24 4 Remedial Mathematics for B. Pharmacy = = 0 ( + 5) = 0 = 0, = 5 Hence the solution set {0, 5} Problem 7: ( + ) ( ) ( ) = 6 The given equation can be written as 6 ( + 0) ( ) = = + ( ) Therefore taking together first and fourth and second and third factors, we get 6 = 4 = Putting = y, put in (A) 6 y (y ) = 4 4y 4y 6 = 0 4y 8y + 4y 6 = 0 y (y 9) + 7 (y 9) = 0 (y 9) (y + 7) = y = or y = 9 Now, when y = Then 9 = 6 4..(A)
25 Equations Reducible to Quadratics Equations 5 9 = = 0 ( ) + ( ) = 0 ( ) ( + ) = 0 = or = Again, When 7 y = Then = = 0 Its discriminant = b 4ac = ( ) 4 7 = 47 < 0 Equation (B) has no real roots. Hence real roots the given equation are, solution set,..(b) Problem 6: Solve Type VI Let A = 5 6 8, B = then A B =. (ii) A B = 5.. (iii) Dividing (iii) by (ii) A + B = 5 then A = = 8
26 6 Remedial Mathematics for B. Pharmacy Squaring the both sides = = = 0 5( 4) + 4( 4) = 0 ( 4) (5 + 4) = 0 4 = 4 or = 5 4 Solution set = 4, 5 Problem = = X+ + 7 X = , 0 Ans:, 0 5 Ans:, Ans:, imaginary Ans: (, ) Ans:, Ans:,,, , 4 4,
27 Equations Reducible to Quadratics Equations 7 7. ( + ) ( + ) ( 5) ( 6) = ( + ) ( + ) ( + ) = y 8 = 0 y. 4. ( ) = Ans: (, 7, ) Ans:, 0, 0 Ans: (, 8, + 5, 5 ) Ans:, Ans: (4) Ans: 5 Ans: (, ) Ans: (, 5) Ans: (4) Solution of Word Problems involving Quadratic Equations We consider such word problems in the eamples given below. Eample : The product of Ramu s age (in years) five years ago with age 9 years later is 5. Find Ramu s present age. Let Ramu s present age = His age 5 years ago = 5
28 8 Remedial Mathematics for B. Pharmacy His age 9 years later = + 9 According to the problem Therefore ( 5) ( + 9) = = 5 i.e., = 0 Solving = 0 ( + 0) 6( + 0) = 0 ( 6) ( + 0) = 0 = 6 or = 0 Since is the present age of Ramu, it cannot be negative. Therefore, we reject the solution = 0. Thus Ramu s present age is 6 years (check Ramu s age 5 years ago = 6 5 = year Ramu s age 9 years later = 5 years Their product = 5, as required in the problem Eample : Find two natural numbers which differ by and whose squares have the sum 7. Let the larger of the two numbers be Then the other number = Sum of their squares = 7 + ( ) = = 0 Dividing both sides by, 54 = = 0 ( 9) + 6( 9) = 0 ( 9) ( + 6) = 0 The solution of this quadratic are = 9 or = 6
29 Equations Reducible to Quadratics Equations 9 Since must be natural number, it cannot be negative and so we reject the solution = 6 Therefore = 9 Therefore, the larger number = 9 The other number = = 6 Therefore, the numbers required are 6 and 9. Check: Sum of their squares = = 7 as required and they differ by. Eample : The product of two successive multiples of 5 is 00. Determine the multipliers. Let the two required numbers be 5 and 5( + ) According to the problem 5 5( + ) = 00 ( + ) = + = 0 ( ) ( + 4) = 0 = 0 = + 4 = 0 = 4 which is rejected as 4 Thus, the numbers are 5, 5( + ) i.e., 5, 0. Eample 4: Divide 6 into two parts such that twice the square of the larger part eceeds the square of the smaller part by 64. Let the larger part be So the smaller part is 6 0 < < 6 According to the problem () = (6 ) + 64 = = = 0 ( + 4) ( 0) = 0 = 4 or = 0
30 0 Remedial Mathematics for B. Pharmacy But = 4 does not satisfy 0 < < 6 so it is rejected. = 0, which is the largest part smaller part = 6 0 = 6 Eample 5: A fast train hours less than a slow train for a journey of 600 km. If the speed of the slow train is 0 km/hr less than that of the fast train, find the speed of two trains. Let the speed of fast train = km/hr Then the speed of slow train = ( 0) km/hr Time taken for journey of 600 km by fast train 600 dis tan ce = hours Time speed And time taken for journey of 600 km by slow train 600 = 0 hours Since the fast train takes hours less than the slow train ( 0) = 600 ( 0) (Multiplying by ( ) = = = 0 ( 50) ( + 40) = 0 = 50 or = 40 Rejecting = 40, as the speed of train cannot be negative. speed of fast train = = 50 km/hr Speed of slow train = 0 = 40 km/hr Eample 6: A motor-boat whose speed is 5 km/hr in still water goes 0 km downstream and comes back in a total of 4 hours 0 minutes. Determine the speed of water. Let km/hr be the speed of water. The speed of motor boat in still water is 5 km/hr
31 Equations Reducible to Quadratics Equations Therefore, its speed downstream is (5 + ) km/hr and the speed upstream is (5 ) km/hr 0 Time taken for going 0 km downstream = hours 5 0 Time taken for going 0 km upstream = hours 5 9 Since the total time is given to be 4 hours 0 minutes i.e., hours hours 0(5 ) 0(5 ) (5 )(5 ) = = 9 (5 ) 00 = 5 = 5 = 5 or = 5 [ 5 as 0] = 5 Hence the speed of water = 5 km/hr. 9 Eample 7: The hypotenuse of a right triangle is m less than twice the shortest side. If the third side is m more than the shortest side, find the sides of the triangle. Let m be the shortest side of the right triangle then hypotenuse = ( )m and third side = ( + )m. using pythagorus theorem ( ) = + ( + ) = = 0 ( ) = 0
32 Remedial Mathematics for B. Pharmacy = 0 or = but = 0 m is in inadmissible as is the length of a side. Thus = m is the length of the shortest side of the triangle. Length of hypotenuse = = = 5m Length of third side = + = 4m Eample 8: The sides (in cm) of a right triangle containing the right angle are 5 and. If the area of the triangle is 60 cm, find the sides of the triangle. Given that area of right triangle ABC = 60 cm Base length of perpendicular = 60 BC AC = 60 5 ( ) = = 0 4 = = 0 ( ) +8( ) = 0 ( ) ( + 8) = 0 8 = or = Since side cannot be negative, So taking = The sides of the triangle are 5 = 5 = 5 cm and = = 8 cm
33 Equations Reducible to Quadratics Equations Eample 9: Find two consecutive positive odd integers whose squares have the sum 90. Let the two consecutive positive odd integers be n + and n +, where n is a whole no. According to the problem (n + ) + (n + ) = 90 (4n + 4n + ) + (4n + n + 9) = 90 8n + 6n 80 = 0 n + n 5 = 0 n + 7n 5n 5 = 0 n(n + 7) 5(n + 7) = 0 (n + 7) (n 5) = 0 n = 7 or n = 5 Since n is a whole number n = 7 is rejected n = 5 Hence, the two consecutive positive odd integers are 5 + and 5 + i.e., and. Eample 0: The Product of two consecutive integers is 7. Find the integers. Let one of the integers be = The second integer = + Now according to the problem ( + ) = 7 + = = = 0 ( + 9) 8( + 9) = 0 ( + 9) = 0 or 8 = 0 = 9 or = 8 Hence the required integers are 8 and 9 or 8 and 9.
34 4 Remedial Mathematics for B. Pharmacy Problem.5. The sum of the squares of there consecutive integers is 50. Find the integers. Ans: 5, 4,, or, 4, 5. A two digit number is such that product of the digits is. When 6 is added to this number the digits interchange their places. Determine the number. Ans: 6. The Perimeter of a rectangle is 8 meter and its area is 400 m. Find the breadth of the rectangle. Ans: 6 m 4. The sides in cm of a right triangle are ( ),, +. Find the sides of the triangle. Ans: cm, 4 cm, 5 cm 5. An integer when added to its square, equals 90. Find all possible values of the integer. Ans: (9, 0)
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