6x^2-3-7x find its zeros and verify the relationship between the zeros and coefficient. please try to solve in copy
Answers
Given polynomial is 6x² - 3 - 7x or 6x² - 7x - 3.
We have to find the zeros of the above polynomial and verify the relationship between the zeros and coefficient.
To find the zeros the given polynomial we have to solve it by splitting the middle term.
Since it is in the form ax² + bx + c = 0. So,
→ 6x² - 7x - 3 = 0
→ 6x² - 9x + 2x - 3 = 0
→ 3x(2x - 3) +1(2x - 3) = 0
→ (2x - 3)(3x + 1) = 0
On comparing we get,
→ x = 3/2, -1/3
Therefore, the zeros of the polynomial are 3/2 and -1/3.
Verification
We know that,
Sum of zeros = -b/a and Product of zeros = c/a
And in a given polynomial; a = 6, b = -7 and c = -3
Now,
Sum of zeros = -b/a
3/2 + (-1/3) = -(-7)/6
(9 - 2)/6 = 7/6
7/6 = 7/6
Product of zeros = c/a
(3/2) × (-1/3) = -3/6
-1/2 = -1/2
Given :-
- 6x² - 7x - 3 = 0 , Find its zeros and verify the relationship between the zeros and coefficient. ?
Solution :-
→ 6x² - 7x - 3 = 0
Splitting The middle Term , we get,
→ 6x² - 9x + 2x - 3 = 0
→ 3x(2x - 3) + 1(2x - 3) = 0
→ (2x - 3)(3x + 1) = 0
Putting Both Equal to Zero now, we get,
→ 2x - 3 = 0
→ 2x = 3
→ x = (3/2)
Or,
→ 3x + 1 = 0
→ 3x = -1
→ x = (-1/3).
______________________________
Now, First Relation is :-
→ Sum of Zeros = - (coefficient of x) /(coefficient of x²)
Putting both values :-
→ (3/2 - 1/3) = -(-7)/6
→ (9 - 2)/6 = 7/6
→ 7/6 = 7/6 ✪✪ Hence Verified. ✪✪
Second Relation :-
→ Product Of Zeros = Constant Term / (coefficient of x²)
Putting both Values :-
→ (3/2) * (-1/3) = (-3) / (6)