Question

find the zeroes of the quadratic polynomial 6x2 + x-2 and verify the relationship between the zeroes and coefficient ​

Answer 1

Given:

The quadratic polynomial 6x2 + x-2

To find:

Find the zeroes of the quadratic polynomial 6x2 + x-2 and verify the relationship between the zeroes and coefficient ​

Solution:

From given, we have,

The quadratic polynomial 6x² + x - 2

In order to find the zeros of this polynomial, we have to equate this polynomial to zero.

So, we get, 6x² + x - 2 = 0

6x² + x - 2 = 0

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

x = 1/2, x = -2/3

The coefficients of the polynomial are:

Sum of zeros = -b/a = - 1/6

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

Compare the zeros of  the  polynomial with the coefficients of the polynomial.

So, the result is,

1/2 ≠ -1/6 and -2/3 ≠ -1/6

The relation is not satisfied.

Answer 2

Question :- Find the zeroes of the quadratic polynomial 6x² + x - 2 and verify the relationship between the zeroes and coefficient ?

Solution :-

Put the given quadratic polynomial equal to 0.

→ 6x² + x - 2 = 0

Now, Splitting The middle term,

→ 6x² + 4x - 3x - 2 = 0

→ 2x(3x + 2) - 1(3x + 2) = 0

→ (2x - 1)(3x + 2) = 0

Putting Both equal to 0 Now, we get,

→ 2x - 1 = 0

→ 2x = 1

→ x = (1/2).

and,

→ (3x + 2) = 0

→ 3x = (-2)

→ x = (-2/3).

Hence, Zeros of the given quadratic polynomial are (1/2) and (-2/3).

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Now,

First Relation is :-

→ Sum of Zeros = - (coefficient of x) /(coefficient of x²)

Putting both values ,

→ (1/2) + (-2/3) = -(1)/6

→ (3 - 4)/6 = (-1/6)

→ (-1/6) = (-1/6) { Hence Verified. }

Second Relation :-

→ Product Of Zeros = Constant Term / (coefficient of x²)

Putting both Values ,

→ (1/2) * (-2/3) = (-2) / 6

→ (-1/3) = (-1/3) { Hence Verified. }

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(Correct Answer.)