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Quadratic Equations : Exercise 4.4 (Mathematics NCERT Class 10th)


                 CLICK HERE to watch the second part

Q.1      Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :
            (i) 2{x^2} - 3x + 5 = 0
            (ii) 3{x^2} - 4\sqrt 3 x + 4 = 0
            (iii) 2{x^2} - 6x + 3 = 0
Sol.       (i) The given equation is 2{x^2} - 3x + 5 = 0
             Here , a = 2, b = – 3 and c = 5
             Therefore, D = {b^2} - 4ac = {\left( { - 3} \right)^2} - 4 \times 2 \times 5 = 9 - 40 = - 31 < 0
             So, the given equation has no real roots.

            (ii) The given equation is 3{x^2} - 4\sqrt 3 x + 4 = 0
            Here a = 3, b = - 4\sqrt 3 \,\,and\,\,c = 4
            Therefore, D = {b^2} - 4ac = {\left( { - 4\sqrt 3 } \right)^2} - 4 \times 3 \times 4 = 48 - 48 = 0
            So, the given equation has real equal roots, given by
            x = {{ - b \pm \sqrt D } \over {2a}} = {{ - \left( { - 4\sqrt 3 } \right) \pm 0} \over {2 \times 2}} = \sqrt 3 .

            (iii) The given equation is 2{x^2} - 6x + 3 = 0
            Here, a = 2, b = – 6 and c = 3
            Therefore,  D = {b^2} - 4ac = {\left( { - 6} \right)^2} - 4 \times 2 \times 3 = 36 - 24 = 12  data-recalc-dims= 0" />
            So, the given equation has real roots , given by
            x = {{ - b \pm \sqrt D } \over {2a}} = {{ - \left( { - 6} \right) \pm \sqrt {12} } \over {2 \times 2}}
             = {{6 \pm 2\sqrt 3 } \over 4} = {{3 \pm \sqrt 3 } \over 2}


Q.2      Find the values of k for each of the following quadratic equations, so that they have two equal roots.
            (i) 2{x^2} + kx + 3 = 0
            (ii) kx\left( {x - 2} \right) + 6 = 0
Sol.      (i) The given equation is 2{x^2} + kx + 3 = 0
            Here, a = 2, b = k and c = 3
            Therefore, D = {b^2} - 4ac = {k^2} - 4 \times 2 \times 3 = {k^2} - 24
            The given equation will have real and equal roots, if
             D = 0  \Rightarrow {k^2} - 24 = 0  \Rightarrow k = \pm \sqrt {24} = \pm 2\sqrt 6
            (ii) The given equation is kx (x – 2) + 6 = 0
             \Rightarrow k{x^2} - 2kx + 6 = 0
            Here a = k , b = – 2k and c = 6
            Therefore, D = {b^2} - 4ac = {\left( { - 2k} \right)^2} - 4 \times k \times 6 = 4{k^2} - 24\,k
            The given equation will have real and equal roots, if D = 0
             \Rightarrow 4{k^2} - 24k = 0  \Rightarrow 4k\left( {k - 6} \right) = 0
             \Rightarrow k = 0\,\,\,or\,\,\,k = 6


Q.3      Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 {m^2} ? If so, find its length and breadth.
Sol.      Let 2x be the length and x be the breadth of a rectangular mango grove.
            Area = (2x) (x) = 800 [Given]
             \Rightarrow {x^2} = 400
             \Rightarrow x = 20 [Since cannot  be negative]
            The value of x is real so design of grove is possible
             Its length = 40 m and breadth = 20 m.


Q.4      Is the following situation possible ? If so, determine their present ages. The sum of the ages of two friend is 20 years. Four year ago, the porudct of their ages in years was 48.
Sol.      Let age of one of the friends = x years
            Then, age of the other friend = 20 – x Four years ago,
            Age of one of the friend = (x – 4) years
            and age of the other friend = (20 – x – 4) years
            = (16 – x) years
            According to condition :
            \left( {x - 4} \right)\left( {16 - x} \right) = 48
             \Rightarrow 16 - {x^2} - 64 + x = 48
             \Rightarrow {x^2} - 20 + 112 = 0
            Here a = 1, b = – 20 and c = 112
            Therefore, D = {b^2} - 4ac = {\left( { - 20} \right)^2} - 4 \times 1 \times 112
             = 400 - 448 = - 48\,\, < 0
            So, the given equation has no real roots.
            Thus , the given situation is not possible.


Q.5      Is it possible to design a rectangular park of perimeter 80 m and area 400 {m^2} ? If so, find its length and breadth.
Sol.      Let length be x metres and breadth be y metres.
            Therefore, Perimeter = 80 m
             \Rightarrow 2\left( {x + y} \right) = 80
             \Rightarrow x + y = 40 ...(1)
            Also, Area = 400\,{m^2}
             \Rightarrow xy = 400
             \Rightarrow x\left( {40 - x} \right) = 400 [Using (1)]
             \Rightarrow 40x - {x^2} = 400
             \Rightarrow {x^2} - 40x + 400 = 0
            Here a = 1, b = – 40 and c = 400
            Therefore,D = {b^2} - 4ac
            = {\left( { - 40} \right)^2} - 4 \times 1 \times 400
            = 1600 - 1600 = 0
           So, the given equation has equal real roots.
           Therefore, Its length and breadth is given by
           {x^2} - 40x + 400 = 0
            \Rightarrow {\left( {x - 20} \right)^2} = 0
            \Rightarrow x = 20,20
           Therefore, Length = 20 m
           Breadth = 20 m
           Therefore, Design is possible.



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