Quadratic Equations : Exercise 4.1 (Mathematics NCERT Class 10th)




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Q.1    Check whether the following are quadratic equations :
          (i) {\left( {x + 1} \right)^2} = 2\left( {x - 3} \right)
          (ii) {x^2} - 2x = \left( { - 2} \right)\left( {3 - x} \right)
          (iii) \left( {x - 2} \right)\left( {x + 1} \right) = \left( {x - 1} \right)\left( {x + 3} \right)
          (iv) \left( {x - 3} \right)\left( {2x + 1} \right) = x\left( {x + 5} \right)
          (v) \left( {2x - 1} \right)\left( {x - 3} \right) = \left( {x + 5} \right)\left( {x - 1} \right)
          (vi) {x^2} + 3x + 1 = {\left( {x - 2} \right)^2}
          (vii) {\left( {x + 2} \right)^3} = 2x\left( {{x^2} - 1} \right)
          (viii) {x^3} - 4{x^2} - x + 1 = {\left( {x - 2} \right)^3}
Sol.    (i) We have {\left( {x + 1} \right)^2} = 2\left( {x - 3} \right)
           \Rightarrow {x^2} + 2x + 1 = 2x - 6
           \Rightarrow {x^2} + 2x + 1 - 2x + 6 = 0
           \Rightarrow {x^2} + 7 = 0
          Clearly, {x^2} + 7 is a quadratic polynomial. So the given equation is a quadratic equation.

         

(ii) We have , {x^2} - 2x = \left( { - 2} \right)\left( {3 - x} \right)
           \Rightarrow {x^2} - 2x + 2\left( {3 - x} \right) = 0
           \Rightarrow {x^2} - 2x + 6 - 2x = 0
           \Rightarrow {x^2} - 4x + 6 = 0
          Clearly, {x^2} - 4x + 6 is a quadratic polynomial. SO, the given equation is a quadratic equation.

         (iii) We have , \left( {x - 2} \right)\left( {x + 1} \right) = \left( {x - 1} \right)\left( {x + 3} \right)
          \Rightarrow {x^2} - x - 2 = {x^2} + 2x - 3
          \Rightarrow {x^2} - x - 2 - {x^2} - 2x + 3 = 0
          \Rightarrow  - 3x + 1 = 0
         Clearly,  - 3x + 1 is linear polynomial  So the given equation is not a quadratic equation.

         (iv) We have, \left( {x - 3} \right)\left( {2x + 1} \right) = x\left( {x + 5} \right)
          \Rightarrow x\left( {2x + 1} \right) - 3\left( {2x + 1} \right) - x\left( {x + 5} \right) = 0
          \Rightarrow 2{x^2} + x - 6x - 3 - {x^2} - 5x = 0
          \Rightarrow {x^2} - 10x - 3 = 0
         Clearly, {x^2} - 10x - 3 is a quadratic polynomial. So, the given equation is a quadratic equation.

         (v) We have, \left( {2x - 1} \right)\left( {x - 3} \right) = \left( {x + 5} \right)\left( {x - 1} \right)
          \Rightarrow \left( {2x - 1} \right)\left( {x - 3} \right) - \left( {x + 5} \right)\left( {x - 1} \right) = 0
          \Rightarrow 2x\left( {x - 3} \right) - 1\left( {x - 3} \right) - x\left( {x - 1} \right) - 5\left( {x - 1} \right) = 0
         \Rightarrow 2{x^2} - 6x - x + 3 - {x^2} + x - 5x + 5 = 0
         \Rightarrow {x^2} - 11x + 8 = 0
        Clearly, {x^2} - 11x + 8 is a quadratic polynomial. So, the given equation is a quadratic equation.

        (vi) We have {x^2} + 3x + 1 = {\left( {x - 2} \right)^2}
         \Rightarrow {x^2} + 3x + 1 - {\left( {x - 2} \right)^2} = 0
         \Rightarrow {x^2} + 3x + 1 - \left( {{x^2} - 4x + 4} \right) = 0
         \Rightarrow {x^2} + 3x + 1 - {x^2} + 4x - 4 = 0
         \Rightarrow 7x - 3 = 0
        Clearly, 7x - 3 is a linear polynomial. So, the given equation is not a quadratic equation.

        (vii) We have, {\left( {x + 2} \right)^3} = 2x\left( {{x^2} - 1} \right)
         \Rightarrow {x^3} + 3{x^2}\left( 2 \right) + 3x{\left( 2 \right)^2} + {\left( 2 \right)^3} = 2{x^3} - 2x
         \Rightarrow {x^3} + 6{x^2} + 12x + 8 - 2{x^3} + 2x = 0
         \Rightarrow  - {x^3} + 6{x^2} + 14x + 8 = 0
        Clearly,  - {x^3} + 6{x^2} + 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, {x^3} - 4{x^2} - x + 1
        = {x^3} + 3{x^2}\left( { - 2} \right) + 3x{\left( { - 2} \right)^2} + {\left( { - 2} \right)^3}
        \Rightarrow {x^3} - 4{x^2} - x + 1 = {x^3} - 6{x^2} + 12x - 8
        \Rightarrow {x^3} - 4{x^2} - x + 1 - {x^3} + 6{x^2} - 12x + 8 = 0
        \Rightarrow 2{x^2} - 13x + 9 = 0
       Clearly, 2{x^2} - 13x + 9 is a quadratic polynomial. So, the given equation is a quadratic equation.


 Q.2       Represent the following situation in the form of quadratic equations.
              (i) The area of a rectangular plot is 528\,{m^2}. 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.
              (ii) The product of two consecutive positive integers is 306. We need to find the integers.
              (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.
              (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.

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 = 528\,{m^2}.
            Since \left( {2x + 1} \right) \times x = 528
             \Rightarrow 2{x^2} + x = 528
             \Rightarrow 2{x^2} + x - 528 = 0,
            Which is the required quadratic equation satisfying  the given conditions.

         

  (ii) Let two consecutive integers be x and x + 1 such that their product = 306.
             \Rightarrow x (x + 1) = 306
             \Rightarrow {x^2} + x - 306 = 0
            Which is the required quadratic equation satisfying  the given conditions.

            (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
             \Rightarrow {x^2} + 32x + 87 = 360
             \Rightarrow {x^2} + 32x + 87 - 360 = 0
             \Rightarrow {x^2} + 32x - 273 = 0
            Therefore, the age of Rohan satisfies the quadratic equation {x^2} + 32x - 273 = 0.

           (iv) Let u km/hr be the speed of the train.
           Then, time taken to cover 480 km  = {{480} \over u}hours
           Time taken to cover 480 km when the speed  is decreased. by 8 km/hr
            = {{480} \over {u - 8}}hours
           It is given that the time to cover 480 km is increased by 3 hours.
           Therefore,  {{480} \over {u - 8}} - {{480} \over u} = 3
            \Rightarrow 480u - 480\left( {u - 8} \right) = 3u\left( {u - 8} \right)
            \Rightarrow 160u - 160u + 1280 = {u^2} - 8u
            \Rightarrow {u^2} - 8u - 1280 = 0
           Therefore, the speed of the train satisfies the quadratic equation {u^2} - 8u - 1280 = 0



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