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Unit-4(Examples)

Example-1 :-  Represent the following situations mathematically:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs 750. We would like to find out the number of toys produced on that day.

Solution :-
(i) Let the number of marbles John had be x. 
  Then the number of marbles Jivanti had = 45 – x. 
  The number of marbles left with John, when he lost 5 marbles = x – 5 
  The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5 = 40 – x

  Therefore, their product = (x – 5) (40 – x) = 40 x – x² – 200 + 5x = – x² + 45x – 200 
  Given that product = 124
  So, – x² + 45x – 200 = 124 
  i.e., – x² + 45x – 324 = 0 
  i.e., x² – 45x + 324 = 0

  Therefore, the number of marbles John had, satisfies the quadratic equation x² – 45x + 324 = 0
    
(ii) Let the number of toys produced on that day be x. 
  Therefore, the cost of production (in rupees) of each toy that day = 55 – x 

  So, the total cost of production (in rupees) that day = x (55 – x) 
  Therefore, x (55 – x) = 750 
  i.e., 55x – x² = 750 
  i.e., – x² + 55x – 750 = 0 
  i.e., x² – 55x + 750 = 0 

  Therefore, the number of toys produced that day satisfies the quadratic equation x² – 55x + 750 = 0 
    

Example-2 :-  Check whether the following are quadratic equations:
(i) (x – 2)² + 1 = 2x – 3
(ii) x(x + 1) + 8 = (x + 2) (x – 2)
(iii) x (2x + 3) = x² + 1
(iv) (x + 2)³ = x³ – 4

Solution :-
(i) L.H.S. 
  (x – 2)2 + 1 = x² – 4x + 4 + 1 = x² – 4x + 5
  Therefore, (x – 2)2 + 1 = 2x – 3 can be rewritten as x² – 4x + 5 = 2x – 3 
  i.e., x² – 6x + 8 = 0 
  It is of the form ax² + bx + c = 0. 

  Therefore, the given equation is a quadratic equation.
    
(ii) Since x(x + 1) + 8 = x² + x + 8 and (x + 2)(x – 2) = x² – 4 
  Therefore, x² + x + 8 = x² – 4 
  i.e., x + 12 = 0 
  It is not of the form ax² + bx + c = 0. 
        
  Therefore, the given equation is not a quadratic equation. 
    
(iii) Here, LHS = x (2x + 3) = 2x² + 3x 
  So, x (2x + 3) = x² + 1 can be rewritten as 2x² + 3x = x² + 1 
  Therefore, we get x² + 3x – 1 = 0 
  It is of the form ax² + bx + c = 0. 

  Therefore, the given equation is a quadratic equation. 
    
(iv) Here, LHS = (x + 2)³ = x³ + 6x² + 12x + 8 
  Therefore, (x + 2)³ = x³ – 4 can be rewritten as x³ + 6x² + 12x + 8 = x³ – 4 
  i.e., 6x² + 12x + 12 = 0 or, x² + 2x + 2 = 0 
  It is of the form ax² + bx + c = 0. 

  Therefore, the given equation is a quadratic equation.
    

Example-3 :-  Find the roots of equation 2x² - 5x + 3 = 0, by factorisation.

Solution :-
  Let us first split the middle term – 5x as –2x –3x [because (–2x) × (–3x) = 6x² = (2x²) × 3]. 
  So, 2x² – 5x + 3 
= 2x² – 2x – 3x + 3 
= 2x (x – 1) –3(x – 1) 
= (2x – 3)(x – 1) 
  Now, 2x² – 5x + 3 = 0 can be rewritten as (2x – 3)(x – 1) = 0. 
  So, the values of x for which 2x² – 5x + 3 = 0 are the same for which (2x – 3)(x – 1) = 0, 
  i.e., either 2x – 3 = 0 or x – 1 = 0.
  Now, 2x – 3 = 0 gives x = 3/2 and x – 1 = 0 gives x = 1.
  So, x = 3/2 and x = 1 are the solutions of the equation.
    

Example-4 :-  Find the roots of the quadratic equation 6x² – x – 2 = 0.

Solution :-
  We have, 
  6x² – x – 2 
= 6x² + 3x – 4x – 2 
= 3x (2x + 1) – 2 (2x + 1) 
= (3x – 2)(2x + 1) 
  The roots of 6x² – x – 2 = 0 are the values of x for which (3x – 2)(2x + 1) = 0 
  Therefore, 3x – 2 = 0 or 2x + 1 = 0,
  i.e., x = 2/3 and x = -1/2

  Therefore, the roots of 6x² – x – 2 = 0 are x = 2/3 and x = -1/2.
    

Example-5 :-  Find the roots of the quadratic equation 3x² - 2√6x + 2 = 0

Solution :-
  We have, 3x² - 2√6x + 2 = 0
  i.e., 3x² - √6x - √6x + 2 = 0
  i.e., √3x(√3x - √2) - √2(√3x - √2) = 0
  i.e., (√3x - √2)(√3x - √2) = 0
  Now, √3x - √2 = 0
       √3x = √2
         x = √2/√3
  Therefore, the roots of 3x² - 2√6x + 2 = 0 are √2/√3 and √2/√3.
    

Example-6 :-  A charity trust decides to build a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth. What should be the length and breadth of the hall? Suppose the breadth of the hall is x metres. Then, its length should be (2x + 1) metres. We can depict this information pictorially as shown in Figure. Find the dimensions of the prayer hall.

Solution :-
  Now, area of the hall = (2x + 1). x m² = (2x² + x) m² 
  So, 2x² + x = 300 (Given) 
  Therefore, 2x² + x – 300 = 0 
  We found that if the breadth of the hall is x m, then x satisfies the equation 2x² + x – 300 = 0. 

  Applying the factorisation method, 
  we write this equation as 2x² – 24x + 25x – 300 = 0 
  2x (x – 12) + 25 (x – 12) = 0 
  i.e., (x – 12)(2x + 25) = 0 

  So, the roots of the given equation are x = 12 or x = – 12.5. 
  Since x is the breadth of the hall, it cannot be negative. 
  Thus, the breadth of the hall is 12 m. 
  Its length = 2x + 1 = 25 m.
    

Example-7 :-  Solve the equation 2x² - 5x + 3 = 0, by the method of completing the square.

Solution :-
  The equation 2x² – 5x + 3 = 0 is the same as x² – 5x/2 + 3/2 = 0
  Now, x² – 5x/2 + 3/2
= (x – 5/4)² - (5/4)² + 3/2
= (x - 5/4)² - (25/16 - 3/2)
= (x - 5/4)² - (25 - 24)/16
= (x - 5/4)² - 1/16

  Therefore, 2x² – 5x + 3 = 0 can be written as (x - 5/4)² - 1/16 = 0
  Now, (x - 5/4)² - 1/16 = 0
  (x - 5/4)² = 1/16 
  x - 5/4 = ± 1/4
  x = 5/4 ± 1/4
  x = 5/4 + 1/4 or x = 5/4 - 1/4
  x = 6/4 or x = 4/4
  x = 3/2 or x = 1
  Therefore, the solutions of the equations are x = 3/2 or x = 1.
    

Example-8 :-  Find the roots of the equation 5x² – 6x – 2 = 0 by the method of completing the square.

Solution :-
  Multiplying the equation throughout by 5, we get 25x² – 30x – 10 = 0 
  This is the same as (5x)² – 2 × (5x) × 3 + 32 – 32 – 10 = 0 
  (5x – 3)² – 9 – 10 = 0 
  (5x – 3)² – 19 = 0 
  (5x – 3)² = 19
  5x - 3 = ± √19
  5x = 3 ± √19
   x = (3 ± √19)/5
   x = (3 + √19)/5 or x = (3 - √19)/5
  Therefore, the roots are x = (3 + √19)/5 or x = (3 - √19)/5.
    

Example-9 :-  Find the roots of 4x² + 3x + 5 = 0 by the method of completing the square.

Solution :-
   4x² + 3x + 5 = 0 is the same as 
   (2x)² + 2 . 2x . 3/4 + (3/4)² - (3/4)² + 5 = 0
   (2x + 3/4)² - 9/16 + 5 = 0
   (2x + 3/4)² - (9/16 - 5) = 0
   (2x + 3/4)² - (9 - 80)/16 = 0
   (2x + 3/4)² + 71/16 = 0
   (2x + 3/4)² = -71/16 < 0 
   
   But (2x + 3/4)² can't be negative for any real value of x. 
   So, there is no real value of x satisfying the given equation. 
   Therefore, the given equation has no real roots.
    

Example-10 :-  The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot. Solve by using the quadratic formula.

Solution :-
  Let the breadth of the plot be x metres. 
  Then the length is (2x + 1) metres. 
  Then we are given that x(2x + 1) = 528, 
  i.e., 2x² + x – 528 = 0. 
  This is of the form ax² + bx + c = 0, where a = 2, b = 1, c = – 528. 
  So, the quadratic formula gives us the solution as
  
  x = 64/4 or x = -66/4
  Since x cannot be negative, being a dimension, the breadth of the 
  plot is 16 metres and hence, the length of the plot is 33m.
    

Example-11 :-  Find two consecutive odd positive integers, sum of whose squares is 290.

Solution :-
  Let the smaller of the two consecutive odd positive integers be x. 
  Then, the second integer will be x + 2. 
  According to the question,
  x² + (x + 2)² = 290
  x² + x² + 4x + 4 = 290  
  2x² + 4x – 286 = 0  
  x² + 2x – 143 = 0 which is a quadratic equation in x. 
  Using the quadratic formula, we get
  
  x = 11 or x = – 13 
  But x is given to be an odd positive integer. 
  Therefore, x ≠ – 13, x = 11. 
  Thus, the two consecutive odd integers are 11 and 13. 
    

Example-12 :-  A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m (see Figure). Find its length and breadth.

Solution :-
  Let the breadth of the rectangular park be x m. 
  So, its length = (x + 3) m. 
  Therefore, the area of the rectangular park = x(x + 3) m² = (x² + 3x) m². 
  Now, base of the isosceles triangle = x m.
  Therefore, its area = 1/2 . x . 12 = 6x m²
  According to our requirements,
  x² + 3x =6 x + 4  
  x² – 3x – 4 = 0 

  Using the quadratic formula, we get
  x = (3 ± √25)/2 = (3 ± 5)/2 
  x = (3 + 5)/2 = 8/2 = 4 or 
  x = (3 - 5)/2 = -2/2 = -1
  But x ≠ – 1. 
  Therefore, x = 4. So, the breadth of the park = 4m and its length will be 7m. 
    

Example-13 :-  Find the roots of the following quadratic equations, if they exist, using the quadratic formula:
(i) 3x² – 5x + 2 = 0
(ii) x² + 4x + 5 = 0
(iii) 2x² – 2√2 x + 1 = 0

Solution :-
(i) 3x² – 5x + 2 = 0 
  Here, a = 3, b =  – 5, c = 2. 
  So, b² – 4ac = 25 – 24 = 1 > 0.
  Therefore, x  = (5 ± √1)/6 = (5 ± 1)/6
  x = (5 + 1)/6 = 6/6 = 1 or
  x = (5 - 1)/6 = 4/6 = 2/3
  So, the roots are 2/3 and 1.
    
(ii) x² + 4x + 5 = 0 
  Here, a = 1, b = 4, c = 5. 
  So, b² – 4ac = 16 – 20 = – 4 < 0.
  Since the square of a real number cannot be negative, therefore D² = b² - 4ac will not have any real value.
  So, there are no real roots for the given equation.
    
(iii) 2x² – 2√2 x + 1 = 0
  Here, a = 2, b = -2√2 , c = 1.
  So, b² – 4ac = 8 – 8 = 0
  Therefore, x = (2√2 ± √0)/4 = 2√2/4 = √2/2 = 1/√2 
  So, the roots are 1/√2 and 1/√2 .
    

Example-14 :-  Find the roots of the following equations:
(i) x + 1/x = 3, x ≠ 0
(ii) 1/x - 1/(x - 2) = 3, x ≠ 0, 2

Solution :-
(i) x + 1/x = 3, x ≠ 0
  Multiplying throughout by x, we get
  x² + 1 = 3x i.e., x² – 3x + 1 = 0, which is a quadratic equation. 
  Here, a = 1, b = – 3, c = 1 
  So, b² – 4ac = 9 – 4 = 5 > 0
  Therefore, x = (3 ± √5)/2
  So, the roots are (3 + √5)/2 and (3 - √5)/2.
    
(ii) 1/x - 1/(x - 2) = 3, x ≠ 0, 2
  As x ≠ 0, 2, multiplying the equation by x (x – 2), 
  we get (x – 2) – x = 3x(x – 2) = 3x² – 6x
  So, the given equation reduces to 3x² – 6x + 2 = 0, which is a quadratic equation.
  Here, a = 3, b = – 6, c = 2. 
  So, b² – 4ac = 36 – 24 = 12 > 0
  Therefore, x = (6 ± √12)/6 = (6 ± 2√3)/6 = (3 ± √3)/3
  So, the roots are (3 + √3)/3 and (3 - √3)/3.
    

Example-15 :-  A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Solution :-
  Let the speed of the stream be x km/h. 
  Therefore, the speed of the boat upstream = (18 – x) km/h and 
  the speed of the boat downstream = (18 + x) km/h.
  The time taken to go upstream =distance/speed = 24/(18 - x)hours
  Similarly, the time taken to go downstream = 24/(18 + x)hours

  According to the question :
  24/(18 - x) - 24/(18 + x) = 1
  24(18 + x) - 24(18 - x) = (18 – x) (18 + x)  
  x² + 48x – 324 = 0
  Using the quadratic formula, we get
  b² - 4ac = (48)² - 4 x 1 x (-324) = 2304 + 1296 = 3600
  x = (-48 ± √3600)/2
  x = (-48 ± 60)/2 = 6 or -54

  Since x is the speed of the stream, it cannot be negative. 
  So, we ignore the root x = – 54. 
  Therefore, x = 6 gives the speed of the stream as 6 km/h.
    

Example-16 :-  Find the discriminant of the quadratic equation 2x² – 4x + 3 = 0, and hence find the nature of its roots.

Solution :-
  The given equation is of the form ax² + bx + c = 0, 
  where a = 2, b = – 4 and c = 3. 
  Therefore, the discriminant (D) = b² – 4ac 
= (– 4)² – (4 × 2 × 3) 
= 16 – 24 
= – 8 < 0 
  So, the given equation has no real roots.
    

Example-17 :-  A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

Solution :-
  
  Let P be the required location of the pole. 
  Let the distance of the pole from the gate B be x m, i.e., BP = x m. 
  Now the difference of the distances of the pole from the two gates = AP – BP (or, BP – AP) = 7 m. 
  Therefore, AP = (x + 7) m.

  Now, AB = 13m, and since AB is a diameter, ∠APB = 90°  
  Therefore, AP² + PB² = AB² (By Pythagoras theorem)  
  (x + 7)² + x² = 13 2 i.e., 
  x² + 14x + 49 + x² = 169 i.e., 
  2x² + 14x – 120 = 0 

  So, the distance ‘x’ of the pole from gate B satisfies the equation x² + 7x – 60 = 0 
  So,  it would be possible to place the pole if this equation has real roots. 
  To see if this is so or not, let us consider its discriminant. 

  The discriminant is b² – 4ac = 7² – 4 × 1 × (– 60) = 289 > 0. 
  So, the given quadratic equation has two real roots, 
  and it is possible to erect the pole on the boundary of the park. 
  Solving the quadratic equation x² + 7x – 60 = 0, by the quadratic formula, we get
  x = (-7 ± √289)/2 = (-7 ± 17)/2
  x = (-7 + 17)/2 = 10/2 = 5 or
  x = (-7 - 17)/2 = -24/2 = -12
  Therefore, x = 5 or – 12. Since x is the distance between the pole and the gate B, it must be positive. 
  Therefore, x = – 12 will have to be ignored. So, x = 5. 
  Thus, the pole has to be erected on the boundary of the park at a distance of 5m from the gate B and 12m from the gate A.
    

Example-18 :-  Find the discriminant of the equation 3x² – 2x + 1/3 = 0 and hence find the nature of its roots. Find them, if they are real.

Solution :-
  We have, 3x² – 2x + 1/3 = 0
  Here a = 3, b = -2 and c = 1/3
  Therefore, discriminant (D) = b² - 4ac
= (-2)² - 4 x 3 x 1/3 = 4 - 4 = 0
  Hence, the given quadratic equation has two equal real roots.
  The roots are -b/2a, -b/2a i.e., 2/6, 2/6 i.e., 1/3, 1/3.
    
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