Exercise 4.1

Q.1    Check whether the following are quadratic equations :          (i) (x+1)2=2(x−3)           (ii) x2−2x=(−2)(3−x)           (iii) (x−2)(x+1)=(x−1)(x+3)           (iv) (x−3)(2x+1)=x(x+5)           (v) (2x−1)(x−3)=(x+5)(x−1)           (vi) x2+3x+1=(x−2)2           (vii) (x+2)3=2x(x2−1)           (viii) x3−4x2−x+1=(x−2)3

Sol.    (i) We have (x+1)2=2(x−3)
          ⇒ x2+2x+1=2x−6           ⇒ x2+2x+1−2x+6=0
          ⇒ x2+7=0           Clearly, x2+7 is a quadratic polynomial. So the given equation is a quadratic equation.
          
(ii) We have , x2−2x=(−2)(3−x)
          ⇒ x2−2x+2(3−x)=0           ⇒ x2−2x+6−2x=0           ⇒ x2−4x+6=0           Clearly, x2−4x+6 is a quadratic polynomial. SO, the given equation is a quadratic equation.

         (iii) We have , (x−2)(x+1)=(x−1)(x+3)
         ⇒ x2−x−2=x2+2x−3
         ⇒ x2−x−2−x2−2x+3=0
         ⇒ −3x+1=0
         Clearly, −3x+1 is linear polynomial  So the given equation is not a quadratic equation.
         (iv) We have, (x−3)(2x+1)=x(x+5)
         ⇒ x(2x+1)−3(2x+1)−x(x+5)=0
         ⇒ 2x2+x−6x−3−x2−5x=0
         ⇒ x2−10x−3=0
         Clearly, x2−10x−3 is a quadratic polynomial. So, the given equation is a quadratic equation.
         (v) We have, (2x−1)(x−3)=(x+5)(x−1)
         ⇒ (2x−1)(x−3)−(x+5)(x−1)=0
         ⇒ 2x(x−3)−1(x−3)−x(x−1)−5(x−1)=0
        ⇒ 2x2−6x−x+3−x2+x−5x+5=0
        ⇒ x2−11x+8=0
        Clearly, x2−11x+8 is a quadratic polynomial. So, the given equation is a quadratic equation.
        (vi) We have x2+3x+1=(x−2)2
        ⇒ x2+3x+1−(x−2)2=0
        ⇒ x2+3x+1−(x2−4x+4)=0
        ⇒ x2+3x+1−x2+4x−4=0
        ⇒ 7x−3=0
        Clearly, 7x−3 is a linear polynomial. So, the given equation is not a quadratic equation.
        (vii) We have, (x+2)3=2x(x2−1)
        ⇒ x3+3x2(2)+3x(2)2+(2)3=2x3−2x
        ⇒ x3+6x2+12x+8−2x3+2x=0
        ⇒ −x3+6x2+14x+8=0
        Clearly, −x3+6x2+14x+8 being a polynomial of degree 3, is not a quadratic polynomial. So the given equation is not a quadratic equation.
       (viii) We have, x3−4x2−x+1
       =x3+3x2(−2)+3x(−2)2+(−2)3        ⇒ x3−4x2−x+1=x3−6x2+12x−8
       ⇒ x3−4x2−x+1−x3+6x2−12x+8=0
       ⇒ 2x2−13x+9=0
       Clearly, 2x2−13x+9 is a quadratic polynomial. So, the given equation is a quadratic equation.

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 Q.2       Represent the following situation in the form of quadratic equations.              (i) The area of a rectangular plot is 528m2 . 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.                          
Sol.      (i) Let the length and breadth of the rectangular plot be 2x + 1 metres and x metres respectively. It is given that its area = 528m2 .            Since (2x+1)×x=528
            ⇒ 2x2+x=528
            ⇒ 2x2+x−528=0 ,            Which is the required quadratic equation satisfying  the given conditions.

           (ii) The product of two consecutive positive integers is 306. We need to find the integers.
  (ii) Let two consecutive integers be x and x + 1 such that their product = 306.
            ⇒ x (x + 1) = 306            ⇒ x2+x−306=0             Which is the required quadratic equation satisfying  the given conditions.

  (iii) Rohan's mother is 26 year older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan's present age.             

            (iii) Let Rohan 's present age be x years. Then,            His mother's age = (x + 26) years.            After 3 years, their respective ages are (x + 3) years and (x + 29) years. ... (1)            It is given that the product of, ages mentioned at (1) is 360            i.e. (x + 3)(x + 29) = 360            ⇒ x2+32x+87=360             ⇒ x2+32x+87−360=0             ⇒ x2+32x−273=0             Therefore, the age of Rohan satisfies the quadratic equation x2+32x−273=0

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train
           (iv) Let u km/hr be the speed of the train.           Then, time taken to cover 480 km =480uhours            Time taken to cover 480 km when the speed  is decreased. by 8 km/hr           =480u−8hours            It is given that the time to cover 480 km is increased by 3 hours.           Therefore, 480u−8−480u=3            ⇒ 480u−480(u−8)=3u(u−8)            ⇒ 160u−160u+1280=u2−8u            ⇒ u2−8u−1280=0            Therefore, the speed of the train satisfies the quadratic equation u2−8u−1280=0