Exercise 2.1

                                                                  EXERCISE 2.1

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 + (iii) 6x²- 3 -7x  (iv) 4u²+ 8u (v) t²-15 (vi) 3x² -x- 4  

Solution 1: (i)  x²- 2x -8  

By splitting the middle term we can write the above equation as

x²-4x+2x-8 = x(x-4)+2(x-4) = (x-4) (x+2)

to find its zero we put (x-4) =0 or (x+2)=0

⇒ x =4 or x= -2

VERIFICATION:

Sum of zeros = 4 +(-2) = 4-2 =2 = - (Coefficient of x)

                                                           (Coefficient of x²)

Product of zeros = 4 x (-2) = -8  =  Constant

                                                  (Coefficient of x²)

(ii)  4S²-4s +1

   

By splitting the middle term we can write the above equation as

4S²-2S-2S-1 = 2S(2S-1) -2S(2S-1)  =  (2S-1) (2S-1).

To find its zeros we put either (2S-1)=0 OR (2S-1) =0

⇒2S-1=0 ⇒ S=1/2     OR 2S-1=0 ⇒ S=1/2

Sum of zeros = 1/2 + 1/2 = 1 =  - (Coefficient of x)                                                                                                                                         (Coefficient of x²)

Product of Zeros = 1/2 X 1/2 = 1/4 = Constant

                                                         (Coefficient of x²)

(iii) 6x²- 3 -7x

Solution: 6x² - 7x -3 = 6x²-9x+2x-3  =  3x(2x-3) +1(2x-3)  = (2x-3)(3x+1)

To find its zeros we put either 

⇒2x-3=0⇒2x=3⇒x=3⁄2    OR  3X+1 = 0 ⇒ 3x=-1 ⇒ x= -1⁄3

VERIFICATION

Sum of zeros = 3⁄2 + -1/3  = (9-2)/6 = 7/6 = -Coefficient of x

                                                                      Coefficient of x²

Product of zeros = 3/2 x -1/3 =-1/2 = Constant

                                                       Coefficient of x²

(iv)   4u²+ 8u

  = 4u(u+2) 

To find its zeros we put either 4u =0 OR (u+2) = 0

⇒ u=0   OR u = -2

VERIFICATION

Sum of zeros = 0 +(- 2) =   - 2 =  -Coefficient of u    

                                                      Coefficient of u²

Product of zeros = 0 x -2 = 0 =   Constant

                                            Coefficient of u²

(v) t²-15

 Solution:  t² - (√15)²  = (t-√15)(t+√15)

To find its zeros we put either  (t-√15)=0 OR (t+√15)=0

⇒t = √15  OR   t = -√15

VERIFICATION:

Sum of zeros = √15 - √15 = 0 = - Coefficient of t

                                                                               Coefficient of t²

Product of zeros  = √15 x -√15  = -15  = Constant 

                                                   Coefficient of t²

(vi) 3x² -x- 4  

Solution:  3x² -x- 4  =  3x² -4x+3x- 4  = x(3x-4) +1(3x-4) = (3x-4)(x+1)

To find its zeros we put 3x-4 =0 OR x+1 =0

⇒3x=4⇒ x = 4⁄3   OR x =-1

VERIFICATION

Sum of zeros = 4/3+( -1) =1/3  = - Coefficent of x² 

                                                      Coefficient of x

Product of zeros =4/3 x -1  = -4/3 = Constant  

                                                      Coefficent of x²

Q.2       Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.            (i) 14,−1                 
            (ii) 2–√,13                      
            (iii) 0,5–√            (iv) 1, 1
            (v) −14,14                                                  

            (vi) 4, 1

Note: Required quadratic polynomial can be obtained by inserting the given values in the formulae   x² - (S)x +P , Where S is sum of zeros and P is product of zeros.

Solution: (i) 14,−1  Here Sum of zeros =  14,and Product of zeros= −1141414

Therefore required quadratic polynomial is x² - (S)x + (P) = x² - (1/4)x + 114

1

(ii) 2–√,13

 Here Sum of zeros = √2 and product of zeros = 1/3

Therefore required quadratic polynomial is x² - (S)x + (P) = x² - √2 x + 1/3

(iii) 0,5–√  Here Sum of zeros = 0  and product of zeros =√5

Therefore required quadratic polynomial is x² - (S)x + (P) = x²-0x+√5 = x²+ √5

(iv) 1, 1

0,5–
  Here Sum of zeros = 1  and product of zeros =1

Therefore required quadratic polynomial is x² - (S)x + (P) = x²-x+1

(v) −14,14

Here Sum of zeros =  14,and Product of zeros= −1/4141414

Therefore required quadratic polynomial is x² - (S)x + (P) = x² - (1/4)x + 1/4

(vi) 4, 1 

  Here Sum of zeros = 4  and product of zeros =1

Therefore required quadratic polynomial is x² - (S)x + (P) = x²-4x+1

                                              EXERCISE 2.2