Question

* Find the zeroes of the quadratic polynomial
6x(square)-3 - 7x and verify the relationship between
the zeroes and coefficients of the polynomial
​

Answer 1

Solution:

The given Equation is f(x) = 6x² - 7x - 3

f(x) = 6x² - 7x - 3

☞ 6x² + 2x - 9x - 3

☞ 2x(3x + 1) - 3(3x + 1)

☞ (2x-3)(3x+1)

•°• Zeroes of Polynomial ;

f(x) = 2x - 3. 0r 3x + 1

☞ 2x - 3 = 0. 0r 3x + 1 = 0

☞ 2x = 3. 0r 3x = - 1

☞ x = 3/2. 0r x = -1/3

Now, Relationship between Zeroes;

a+b = - b/a

☞ 3/2 +(-1/3)

☞ 3/2 - 1/3

☞ 9 - 2 / 6

☞ 7/6 ( It is Equal to Sum of Zeroes )

Case : II

☞ aß = c/a

☞ 3/2 × -1/3

☞ -1/2 { It is Equal to Product of Zeroes }

Therefore, It's Proved.

Answer 2

Given polynomial :-

6x^2 -3 -7x

let's rearrange the terms in it as the general quadratic polynomial should be in the form of ax^2 + bx + c

→ 6x^2 -7x -3

To find zeroes We have to do factorization.

Factorization in brief :-

product of First and last terms

6 × (-3) = - 18

» List out the factors of 18

1× 18 = 18

2× 9 = 18.....

» when we add or subtract the factors we must get the middle term

2 - 9 = -7

Now

6x^2 +2x -9x -3 =0

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

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

◆ 2x - 3 = 0

2x = 3

x = 3 /2

◆ 3x + 1 = 0

3x = -1

x = -1/3

So the two zeroes of the polynmial are 3/2 and -1/3

Lets verify the relationship between zeroes and coefficients

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

sum of the zeroes :-

3/2 + (-1/3 )

3/2 - 1/3 = 9 -2/6 = 7/6 ----(1)

we know that sum of the zeroes is -b/a

-(-7)/6 = 7 /6 ------(2)

product of the zeroes :-

3/2 × -1/3 = -3/6---------(3)

we Know that product of zeroes is c/a

-3/6----------(4)

•°• (1) = (2)

(3) = (4)

Hence the relationship between zeroes and coefficients is verified