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Polynomials : Exercise - 2.2 (Mathematics NCERT Class 10th)


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Q.1       Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients :
             (i) {x^2} - 2x - 8                      
             (ii) 4{s^2} - 4s + 1
             (iii) 6{x^2} - 3 - 7x                                         
             (iv) 4{u^2} + 8u
             (v) {t^2} - 15                                                    
             (vi) 3{x^2} - x - 4

Sol.       (i) We have, {x^2} - 2x - 8  = {x^2} + 2x - 4x - 8
                                                    = x\left( {x + 2} \right) - 4\left( {x + 2} \right)
                                                    = \left( {x + 2} \right)\left( {x - 4} \right)

                  The value of {x^2} - 2x - 8 is zero when the value of (x + 2) (x – 4) is zero, i.e.,
                  when x + 2 = 0 or x – 4 = 0 , i.e., when x = – 2 or x = 4.

                   So, The zeroes of {x^2} - 2x - 8 are – 2 and 4.
                  Therefore , sum of the zeroes = (– 2) + 4 = 2
                                                              = {{ - Coefficient\,of\,x} \over {Coefficient\,of\,{x^2}}}
                  and product of zeroes = (– 2) (4) = – 8  = {{ - 8} \over 1}
                                                   ={{Cons\tan t\,term} \over {Coefficient\,of\,{x^2}}}

            (ii) We have, 4{s^2} - 4s + 1  = 4{s^2} - 2s - 2s + 1
                                                     = 2s\left( {2s - 1} \right) - 1\left( {2s - 1} \right)
                                                     = \left( {2s - 1} \right)\left( {2s - 1} \right)
                 The value of 4{s^2} - 4s + 1 is zero when the value of  
                 (2s – 1) (2s – 1) is zero, i.e., when 2s – 1 = 0 or 2s – 1 = 0,
                 i.e., when s = {1 \over 2}\,or\,s = {1 \over 2}.
                 So, The zeroes of 4{s^2} - 4s + 1\,are\,{1 \over 2}\,and\,{1 \over 2}
                 Therefore, sum of the zeroes  = {1 \over 2} + {1 \over 2} = 1
                                                            = {{-Coefficient\,of\,s} \over {Coefficient\,of\,{s^2}}}
                 and product of zeroes  = \left( {{1 \over 2}} \right)\left( {{1 \over 2}} \right) = {1 \over 4}
                                                  = {{Constant\,term} \over {Coefficient \,of\,{s^2}}}
           (iii) We have, 6{x^2} - 3 - 7x6{x^2} - 7x - 3
                                                    = 6{x^2} - 9x + 2x - 3
                                                    = 3x\left( {2x - 3} \right) + 1\left( {2x - 3} \right)
                                                    = \left( {3x + 1} \right)\left( {2x - 3} \right)
                 The value of 6{x^2} - 3 - 7x is zero when the value of (3x + 1) (2x – 3) is zero, i.e., when 3x + 1 = 0 or 2x – 3 = 0, i.e, when x = - {1 \over 3} or x = {3 \over 2}
                So, The zeroes of 6{x^2} - 3 - 7x\,\,\,are\,\, - {1 \over 3}\,and\,\,{3 \over 2}
                Therefore, sum of the zeroes  = - {1 \over 3} + {3 \over 2} = {7 \over 6}
                                                            = {{ - Coefficient\,of\,x} \over {Coefficient\,of\,{x^2}}}
                 and product of zeroes  = \left( { - {1 \over 3}} \right)\left( {{3 \over 2}} \right) = {{ - 1} \over 2}
                                                  = {{Constant\,term} \over {Coefficient \,of\,{x^2}}}
           (iv)  We have, 4{u^2} + 8u = 4u (u + 2) 
                  The value of 4{u^2} + 8u is zero when the value of 4u(u + 2) is zero, i.e., when u = 0 or u + 2 = 0, i.e., when u = 0 or u = – 2.
                  So, The zeroes of 4{u^2} + 8u and 0 and – 2 
                  Therefore, sum of the zeroes = 0 + (– 2) = – 2 
                                                            = {{Coefficent\,of\,u} \over {Coefficient\,of\,{u^2}}}
                  and , product of zeroes = (0) (–2) = 0 
                                                     = {{Cons\tan t\,term} \over {Coefficient\,of\,{u^2}}}
            (v) We have {t^2} - 15  = \left( {t - \sqrt {15} } \right)\left( {t + \sqrt {15} } \right)
                 The value of {t^2} - 15 is zero when the value of \left( {t - \sqrt {15} } \right)\left( {t + \sqrt {15} } \right) is zero,
                 i.e., when 
t - \sqrt {15} = 0\,\,or\,\,t + \sqrt {15} = 0 i.e., when t = \sqrt {15} \,\,or\,\,t = - \sqrt {15}
                 So, The zeroes of {t^2} - 15\,are\,\sqrt {15} \,\,and\,\, - \sqrt {15}
                 Therefore , sum of the zeroes = \sqrt {15} + \left( { - \sqrt {15} } \right) = 0
                  = {{ - Coefficient\,of\,t} \over {Coefficient\,of\,{t^2}}}
                 and, product of the zeroes = \left( {\sqrt {15} } \right)\left( { - \sqrt {15} } \right) = - 15
                                                     = {{Cons\tan t\,term} \over {Coefficient\,of\,{t^2}}}  
           (vi) We have, 3{x^2} - x - 4 =   3{x^2} + 3x - 4x - 4
                                                   = 3x\left( {x + 1} \right) - 4\left( {x + 1} \right)
                                                   = \left( {x + 1} \right)\left( {3x - 4} \right)
                 The value of 3{x^2} - x - 4 is zero when the value of (x + 1) (3x – 4) is zero, i.e., when x + 1 = 0 or 3x – 4 = 0, i.e., when x = – 1 or  x = {4 \over 3}.
                 So, The zeroes of  3{x^2}-x-4\,are\,-1\,and\,{4 \over 3}
                 Therefore , sum of the zeroes  = - 1 + {4 \over 3} = {{ - 3 + 4} \over 3}
                                                               = {1 \over 3} = {{Coefficient\,of\,x} \over {Coefficient\,of\,{x^2}}}
                 and, product of the zeroes  = \left( { - 1} \right)\left( {{4 \over 3}} \right) = - {4 \over 3}
                                                         = {{Constant\,term} \over {Coefficient\,of\,{x^2}}}


Q.2       Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
            (i) {1 \over 4}, - 1                 
            (ii) \sqrt 2 ,{1 \over 3}                      
            (iii) 0,\sqrt 5
            (iv) 1, 1
            (v)  - {1 \over 4},{1 \over 4}                                                  

            (vi) 4, 1
Sol.      (i) Let the polynomial be a{x^2} + bx + c , and its zeroes be  \alpha \,\,and\,\,\beta . Then , 
                                                    \alpha \, + \,\,\beta = {1 \over 4} = {{ - b} \over a}
                  and,                            \alpha \,\beta = - 1 = {{ - 4} \over 4} = {c \over a}
                 If a = 4,  then  b = – 1 and c = – 4.
                 Therefore, one quadratic polynomial which fits the given conditions is 4{x^2} - x - 4.

          (ii) Let the polynomial be a{x^2} + bx + c, and its zeroes be \alpha \,and\,\beta . Then,
                                                 \alpha \, + \,\beta = \sqrt 2 = {{3\sqrt 2 } \over 3} = {{ - b} \over a}
                and                            \alpha \,\beta = {1 \over 3} = {c \over a}
                If a = 3, then b  = 3\sqrt 2 \,and\,c = 1
                So, One quadratic polynomial which fits the given conditions is 3{x^2} - 3\sqrt 2 \,x + 1.

           (iii) Let the polynomial  be a{x^2} + bx + c, and its zeroes be \alpha \,and\,\beta . Then, 

                                                  \alpha + \beta = 0 = {0 \over 1} = {{ - b} \over a}
                 and                           \alpha \beta = \sqrt 5 = {{\sqrt 5 } \over 1} = {c \over a}
                 If  a = 1, then b = 0 and c = {\sqrt 5 }
                 So, one quadratic polynomial which fits the given conditions is {x^2} - 0.x + \sqrt 5 ,\,i.e.,\,{x^2} + \sqrt 5 .

          (iv) Let the polynomial be a{x^2} + bx + c and its zeroes be \alpha \,and\,\beta . Then,
                                                 \alpha \, + \,\beta = 1
                                                  = {{ - \left( { - 1} \right)} \over 1} = {{ - b} \over a}
a
nd                               \alpha \beta = 1  = {1 \over 1} = {c \over a}
                If  a = 1, then b = – 1 and c = 1.
                So, one quadratic polynomial which fits the given conditions is {x^2} - x + 1.

            (v) Let the polynomial be a{x^2} + bx + c and its zeroes be \alpha \,and\,\beta . Then
                                                   \alpha + \beta = - {1 \over 4} = {{ - b} \over a} 
                  and                               \alpha \beta = {1 \over 4} = {c \over a}
                  If  a = 4 then b = – 1 and c = 1. 
                  So, one quadratic polynomial which fits the given conditions is 4{x^2} - x + 1.

            (vi) Let the polynomial be a{x^2} + bx + c and its zeroes be \alpha \,and\,\beta . Then, 
                                                   \alpha \, + \,\beta = 4 = {{ - b} \over a}

                                                   \alpha \,\,\beta = 1 = {1 \over 1} = {c \over a}
                  If a = 1, then b = – 4 and c = 1
                  Therefore, one quadratic polynomial which fits the given conditions is {x^2} - 4x + 1.



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