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Exercise - 4.4

Question-1 :-  Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x² – 3x + 5 = 0
(ii) 3x² – 4√3 x + 4 = 0
(iii) 2x² – 6x + 3 = 0

Solution :-
  We know that for a quadratic equation ax² + bx + c = 0, discriminant is b² − 4ac.
(A) If b² − 4ac > 0 → two distinct real roots 
(B) If b² − 4ac = 0 → two equal real roots 
(C) If b² − 4ac < 0 → no real roots
    
(i) 2x² – 3x + 5 = 0 
  Comparing this equation with ax² + bx + c = 0, we obtain a = 2, b = −3, c = 5
  Discriminant = b² − 4ac = (−3)² − 4 x (2) x (5) = 9 − 40 = −31
  As b² − 4ac < 0,
  Therefore, no real root is possible for the given equation.
    
(ii) 3x² – 4√3x + 4 = 0
  Comparing this equation with ax² + bx + c = 0, we obtain a = 3, b = – 4√3, c = 4
  Discriminant = b² − 4ac = (– 4√3)² − 4 x (3) x (4) = 48 − 48 = 0
  As b² − 4ac = 0
  Therefore, real roots exist for the given equation and they are equal to each other.
  x = -b/2a and x = -b/2a
  x = 4√3/6 = 2√3/3 = 2/√3
  Therefore, the roots are 2/√3	and 2/√3.
    
(iii) 2x² – 6x + 3 = 0 
  Comparing this equation with ax² + bx + c = 0, we obtain a = 2, b = –6, c = 3
  Discriminant = b² − 4ac = (–6)² − 4 x (2) x (3) = 36 − 24 = 12
  As b² − 4ac > 0
  Therefore, distinct real roots exist for this equation as follows.
  x = (6 ± √12)/4 = (6 ± 2√3)/4 = (3 ± √3)/2 
  x = (3 + √3)/2 or x = (3 - √3)/2
  Therefore, the roots are (3 + √3)/2 and (3 - √3)/2.
    

Question-2 :-  Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0

Solution :-
  We know that if an equation ax² + bx + c = 0 has two equal roots, its discriminant (b² − 4ac) will be 0. 

(i) 2x² + kx + 3 = 0
  Comparing equation with ax² + bx + c = 0, we obtain a = 2, b = k, c = 3
  Discriminant = b² − 4ac = (k)² − 4(2) (3) = k² − 24
  For equal roots,
  Discriminant = 0 
  k² − 24 = 0
  k² = 24
  k = ± √24 
  k = ± 2√6
    
(ii) kx (x – 2) + 6 = 0
  kx² − 2kx + 6 = 0
  Comparing this equation with ax² + bx + c = 0, we obtain a = k, b = −2k, c = 6
  Discriminant = b² − 4ac = (− 2k)² − 4 (k) (6) = 4k² − 24k
  For equal roots, b² − 4ac = 0 
  4k² − 24k = 0 
  4k (k − 6) = 0
  Either 4k = 0 or k = 6 = 0 
  k = 0 or k = 6
  However, if k = 0, then the equation will not have the terms ‘x²’ and ‘x’.
  Therefore, if this equation has two equal roots, k should be 6 only. 
    

Question-3 :-  Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.

Solution :-
  Let the breadth of mango grove be x.
  Length of mango grove will be 2x.
  Area of mango grove = (2x) (x) = 2x²

  According to Question : 
  2x² = 800
  x² = 400
  x² - 400 = 0
  Comparing this equation with ax² + bx + c = 0, we obtain
  Discriminant = b2 − 4ac = (0)2 − 4 × (1) × (− 400) = 1600 Here, b2 − 4ac > 0
  Therefore, the equation will have real roots. 
  And hence, the desired rectangular mango grove can be designed.
  x = ± 20

  However, length cannot be negative.
  Therefore, breadth of mango grove = 20 m
  Length of mango grove = 2 × 20 = 40 m 
    

Question-4 :-  Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Solution :-
  Let the age of one friend be x years.
  Age of the other friend will be (20 − x) years. 
  4 years ago, age of 1st friend = (x − 4) years
  And, age of 2nd friend = (20 − x − 4) = (16 − x) years

  According to Question :
  (x − 4) (16 − x) = 48 
  16x − 64 − x² + 4x = 48 
  − x² + 20x − 112 = 0
  x² − 20x + 112 = 0
  Comparing this equation with ax² + bx + c = 0, we obtain a = 1, b = −20, c = 112
  Discriminant = b² − 4ac = (− 20)² − 4 (1) (112) = 400 − 448 = −48
  As b² − 4ac < 0,
  Therefore, no real root is possible for this equation and hence, this situation is not possible.
    

Question-5 :-  Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.

Solution :-
  Let the length and breadth of the park be l and b. 
  Perimeter = 2 (l + b) = 80
  l + b = 40 Or, b = 40 − l
  Area = l × b = l (40 − l) = 40l − l2 
  40l − l2 = 400
  l² − 40l + 400 = 0
  Comparing this equation with al² + bl + c = 0, we obtain a = 1, b = −40, c = 400
  Discriminate = b² − 4ac = (− 40)² − 4 (1) (400) = 1600 − 1600 = 0
  As b² − 4ac = 0,
  Therefore, this equation has equal real roots. 
  And hence, this situation is possible.

  Root of this equation,
  l = -b/2a
  l = -(40)/2 = -20
  Therefore, length of park, l = 20 m
  And breadth of park, b = 40 − l = 40 − 20 = 20 m
    
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