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

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) t² – 15                   (vi) 3x² – x – 4

          

To factorize the value we have to find two value which

sum is equal to, b =-2

 and product is , a*c  = 1*-8 = -8

2 and -4 are required values which

sum is 2-4     = - 2

product is 2*-4 = - 8

So we can write middle term -2 x  = 2 x – 4 x 

We get

           

x (x +2) -4(x+2) = 0

(x+2)(x-4)           = 0

First zero

X+2                 = 0

X                     = - 2

Second zero

x-4                   = 0

x                      = 4

sum of zero              -2 + 4 = 2

product of zero         2*-4     = -8

For a equation ax²+ bx  + c  = 0 , if  zero are α and β ,

Plug the values of a , b and c we get

Sum  of zeros                       -b/a     = -(-2/1) = 2

Product of zeros                    c/a      =-8/1      = -8 

hence we have verified that

(ii) 4s² – 4s + 1

Factorize the equation

  

Compare the equation with as²  + bs  + c  = 0

We get

a = 4 ,b=-4 c=1

To factorize the value we have to find two value which

sum is equal to         b          =-4

product is                   a*c      = 4

-2 and -2 are required values which

sum is            –  2   – 2         =  - 4   

product is       - 2 * - 2            =    4

So we can write middle term         - 4 s   = - 2 s  – 2 s   

We get

2s (2s -1) -1(2s-1) = 0

(2s-1)(2s-1)         = 0

Solve for first zero

2s-1    =0

2s        = 1

s          = ½

Solve for second zero

2s-1    =0

2s        = 1

S         = ½

sum of zeros    1 /2 + 1/2   =2/2 =  1

product of zeros   ½* ½     = 1/4

Sum of zero = -b/a = -(-4/4) = 1

Product of zero= c/a =  1/4 = 1/4

hence we have verified that 

(iii) 6x² – 3 – 7x

Factorize the equation

  

Compare the equation with ax²  + bx  + c  = 0

We get

a = 6 ,b=-7 c= -3

To factorize the value we have to find two value which

sum is equal to         b =-7

product is       a*c  = 6*-3      = -18

2 and -9 are required values which

sum is             2 – 9     = - 7

product is 2 * - 9= - 18

So we can write middle term as

 -7x      = 2x – 9x 

We get

 

2x (3x +1) -3(3x +1) = 0

(3x+1)(2x-3)   = 0

3x+1 = 0 , 2x-3 = 0

Solve for first zero

3x = - 1

x  = -1/3

Solve for second zero

2x-3 = 0

2x    = 3

X      =  3/2

   

Plug the values of a , b and c we get

Sum  of zeros     -b/a     = -(-7/6)      = 7/6

Product of zero    c/a     =  -3/6         = -1/2

hence we have verified that  

(iv) 4u² + 8u

Factorize the equation

Compare the equation with au²  + bu + c  = 0

We get

a = 4 ,b=-8 c= 0

To factorize the value  we can take 4u common there

4u(u+2)= 0

First zero

4u        = 0

u          = 0  

second zero

u+2      = 0

u          = - 2  

sum of zero       0 - 2        = - 2

product of zero  0 * ( - 2 ) = 0

Sum  of zero , -b/a   = -(8/4)    = -2

Product of zero,  c/a =  0/4      =  0

hence we have verified that  

 

(vi) 3x² – x – 4

 Factorize the equation

  

Compare the equation with

ax²  + bx  + c  = 0

We get

a = 3 ,b=-1 c= -4

to factorize the value we have to find two value which

sum is equal to  b    =-1

product is           a*c  = 3*-4  = -12

3 and -4 are required values which

Sum is           3 – 4   = - 1

Product is      3 *- 4   = - 12

So we can write middle term  -x  = 3x – 4x 

We get

3x (x +1) -4(x +1) = 0

(x + 1)( 3x - 4) = 0

x+1 = 0 , 3x-4   = 0

Solve for first zero

x  = - 1

solve for second zero

3x-4 = 0

3x = 4

x = 4/3

  

Sum of zeros  -b/a   = -(-1/3)= 1/3

Product of zero c/a   =  -4/3  = 4/3

hence we have verified that