Class 10 Mathematics - Chapter Polynomials NCERT Solutions | Find the zeroes of the following quadrat

Welcome to the NCERT Solutions for Class 10th Mathematics - Chapter Polynomials. This page offers a step-by-step solution to the specific question from Excercise 2 , Question 1: find the zeroes of the following quadratic polynom....
Question 1

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x2 – 2x – 8 (ii) 4s2 – 4s + 1 (iii) 6x2 – 3 – 7x (iv) 4u2 + 8u (v) t– 15 (vi) 3x– x – 4

Answer

(i)   x2 – 2x – 8

                 = x – 4x + 2x – 8                  

                = x(x – 4) + 2(x – 4)

                 = (x + 2) (x – 4)

                The value of x2 – 2x – 8 is zero if (x + 2) = 0 and (x – 4) = 0

                      x = -2 or   x = 4

                       Sum of zeroes = (-2 + 4) = 2 = - coefficient of x

                                                                        coefficient of x2      

                       Product of zeroes = (-2) × 4 = -8 = Constant term

                                                                             coefficient of x2

(ii)          4s2 – 4s + 1

                          = 4s2 – 2s – 2s + 1

                        = 2s (2s – 1) – 1 (2s – 1)

                         = ( 2s – 1 )  ( 2s – 1 )      

                     The value of 4s2 – 4s + 1 is zero , if (2s-1) = 0 and (2s-1 ) = 0

                       s = 1/2 , 1/2

                       Sum of zeroes = (1/2 + 1/2) = 1   - coefficient of x

                                                                            coefficient of x2

                      Product of zeroes =1/2 × 1/2 = 1/4 =  constant term

                                                                                coefficient of x2             

(iii)    6x2 –7x – 3  

                        = 6x – 9x + 2x – 3                     

                     = 3x (2x – 3) + 1(2x – 3)                        

                     = (3x + 1) (2x – 3)                

                     The value of  6x2 –7x – 3 is zero, if (3x + 1) = 0 and (2x – 3) = 0

                            X = -1 /3 , 3/2

                            Sum of zeroes = ( -1/3 + 3/2) = 7/6 =  - coefficient of x

                                                                                        coefficient of x2

                           Product of zeroes = -1/3 × 3/2 = -3/2 =  constant term

                                                                                        coefficient of x2            

(iv)        4u2+8u

                       4u(u+2)

                       The value of 4u2+8u is zero, if 4u = 0 and (u+2) =0

                        u   = 0,  - 2

                       Sum of zeroes = ( 0+ (-2)) = -2 =  - coefficient of x

                                                                               coefficient of x2

                         Product of zeroes = (-2) × 0 = 0 =   constant term

                                                                                 coefficient of x2            

(v)

solution

(vi) 

3x2–x–4

                            3x – 4x + 3x – 4 

                      =  x (3x – 4) + 1 (3x – 4)

                       The value of 3x – x + 4 is zero, if (3x – 4) = 0 and (x + 1) = 0

                             Sum of zeroes = [4/3 + ( -1)] = 1/3 = - coefficient of x

                                                           coefficient of x2

                          Product of zeroes = (-1) × 4/3 = -4/3 = constant term

                                                                                          coefficient of x2           

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