Math, asked by trishastar313, 7 months ago

Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and their coefficients :​

Attachments:

Answers

Answered by itsbiswaa
0

Answer:

Factorize the equation, we get (x+2)(x−4)

So, the value of x  

2

−2x−8 is zero when x+2=0,x−4=0, i.e., when x=−2 or x=4.

Therefore, the zeros of x  

2

−2x−8 are -2 and 4.

Now,

⇒Sum of zeroes =−2+4=2=−  

1

2

​  

=−  

Coefficient  of  x  

2

 

Coefficient  of  x

​  

 

⇒Product of zeros =(−2)×(4)=−8 =  

1

−8

​  

=  

Coefficient  of  x  

2

 

Constant  term

​  

 

(ii) 4s  

2

−4s+1

Factorize the equation, we get(2s−1)(2s−1)

So, the value of 4s  

2

−4s+1 is zero when 2s−1=0,2s−1=0, i.e., when s=  

2

1

​  

 or s=  

2

1

​  

.

Therefore, the zeros of 4s  

2

−4s+1 are  

2

1

​  

 and  

2

1

​  

.

Now,

⇒Sum of zeroes =  

2

1

​  

+  

2

1

​  

=1=−  

4

−4

​  

=−  

Coefficient  of  s  

2

 

Coefficient  of  s

​  

 

⇒Product of zeros =  

2

1

​  

×  

2

1

​  

=  

4

1

​  

=  

4

1

​  

=  

Coefficient  of  s  

2

 

Constant  term

​  

 

(iii) 6x  

2

−3−7x  

Factorize the equation, we get (3x+1)(2x−3)

So, the value of 6x  

2

−3−7x is zero when 3x+1=0,2x−3=0, i.e., when x=−  

3

1

​  

 or x=  

2

3

​  

.

Therefore, the zeros of 6x  

2

−3−7x are −  

3

1

​  

 and  

2

3

​  

.

Now,

⇒Sum of zeroes = −  

3

1

​  

+  

2

3

​  

=  

6

7

​  

=−  

6

−7

​  

=−  

Coefficient  of  x  

2

 

Coefficient  of  x

​  

 

⇒Product of zeros = −  

3

1

​  

×  

2

3

​  

=−  

2

1

​  

=  

6

−3

​  

=  

2

−1

​  

=  

Coefficient  of  x  

2

 

Constant  term

​  

 

(iv) 4u  

2

+8u  

Factorize the equation, we get 4u(u+2)

So, the value of 4u  

2

+8u is zero when 4u=0,u+2=0, i.e., when u=0 or u=−2.

Therefore, the zeros of 4u  

2

+8u are 0 and −2.

Now,

⇒Sum of zeroes = 0−2=−2=−  

4

8

​  

=−2=−  

Coefficient  of  u  

2

 

Coefficient  of  u

​  

 

⇒Product of zeros = −0x−2=0=  

4

0

​  

=0=  

Coefficient  of  u  

2

 

Constant  term

​  

 

(v) t  

2

−15

Factorize the equation, we get t=±  

15

​  

 

So, the value of t  

2

−15 is zero when t+  

15

​  

=0,t−  

15

​  

=0, i.e., when t=  

15

​  

 or t=−  

15

​  

.

Therefore, the zeros of t  

2

−15 are ±  

15

​  

.

Now,

⇒Sum of zeroes =  

15

​  

−  

15

​  

=0=−  

1

0

​  

=0=−  

Coefficient  of  t  

2

 

Coefficient  of  t

​  

 

⇒Product of zeros =  

15

​  

×  

−15

​  

=−15=  

1

−15

​  

=  

Coefficient  of  t  

2

 

Constant  term

​  

 

(vi) 3x  

2

−x−4

Factorize the equation, we get(x+1)(3x−4)

So, the value of 3x  

2

−x−4 is zero when x + 1 = 0, 3x - 4 = 0, i.e., when x = -1 or x =  

3

4

​  

.

Therefore, the zeros of 3x  

2

−x−4 are -1 and  

3

4

​  

.

Now,

⇒Sum of zeroes = −1+  

3

4

​  

=  

3

1

​  

=−  

3

−1

​  

=−  

Coefficient  of  x  

2

 

Coefficient  of  x

​  

 

⇒Product of zeros = −1×  

3

4

​  

=−  

3

4

​  

=  

3

−4

​  

=  

Coefficient  of  x  

2

 

Constant  term

Step-by-step explanation:

Similar questions