Find the zeros of the following quadratic polynomials and verify the basic relationships between the zeros and the coefficients.

6x² – 3 – 7x

Let f(x) = 6x² – 3 – 7x

Arranging equation in proper form.

Now, f(x) = 6x² – 7x – 3

To find out zeros of the given polynomial.

We put f(x) = 0

⇒ 6x² – 7x – 3 = 0

To find out roots of this polynomial we use splitting of middle term method.

According to this method we need to find two numbers whose sum is – 7 and product is – 18.

∴ 6x² – (9 – 2)x – 3 = 0

⇒6x² – 9x + 2x – 3 = 0

⇒ 3x(2x – 3) + 1(2x – 3) = 0

⇒ (3x + 1)(2x – 3) = 0

∴ 3x + 1 = 0

∴ x = .

Again, 2x – 3 = 0

∴ x =

⇒ Our zeros are α = and β = .

⇒ sum of zeros = α + β = +

⇒ sum of zeros = α + β =

⇒ Product of zeros = αβ = .

Now, Comparing f(x) = 6x² – 7x – 3 with standard equation ax² + bx + c.

We get, a = 6, b = – 7 and c = – 3.

We can verify,

⇒ Sum of zeros =

i.e. α + β =

∴ α + β =

⇒ Product of zeros =

αβ =

Hence, relationship between zeros and coefficient is verified.