Question

1) Find the zeros of the polynomial 6x square - 7x - 3 and verify the relationship the zeros and the coefficients?

Answer 1

$\sf\small\underline\red{Given:-}[\tex]</p><p></p><p>f(x) = 6x2 – 7x – 3</p><p></p><p></p><p>[tex]\sf\small\underline\red{find:-}</p><p>$

he zeros of

Let us put f(x) = 0

⇒ 6x2 – 7x – 3 = 0

⇒ 6x2 – 9x + 2x – 3 = 0

⇒ 3x(2x – 3) + 1(2x – 3) = 0

⇒ (2x – 3)(3x + 1) = 0

⇒ 2x – 3 = 0

x = 3/2

⇒ 3x + 1 = 0

⇒ x = -1/3

It gives us 2 zeros, for x = 3/2 and x = -1/3

Hence, the zeros of the quadratic equation are 3/2 and -1/3.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

3/2 + (-1/3) = – (-7) / 6 7/6 = 7/6

Product of roots = constant / coefficient of x2

3/2 x (-1/3) = (-3) / 6 -1/2 = -1/2

$\sf\small\underline\red{note:-}</p><p>$

Therefore, the relationship between zeros and their coefficients is verified.