Question

find the zeroes of the quadratic polynomial 7x square-x-6 and verify the relationship between the zeroes and the Offieient​

Answer 1

hence zeroes of the polynomialare -6 and 1

let -6 be alpha and 1 be beta

freind your question has some mistake when i verified it i did not get alpha + beta = -b/a

Answer 2

Answer:-

The zeroes of the polynomial 7x² - x - 6 =

→ $\bf{x = 1}$

→ $\bf{x = \dfrac{-6}{7}}$

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• Given-

$7x^2 - x - 6$

• To Find:-

Zeroes of the Polynomial

• Solution:-

→ $7x^2 - x - 6 = 0$

• Splitting the middle term:-

→ $7x^2 - 7x + 6x - 6 = 0$

→ $7x(x - 1) + 6(x - 1) = 0$

→ $(x-1) (7x+6) = 0$

•Now, Taking (x - 1):-

→ $x - 1 = 0$

→ $\bf{x = 1}$

• Taking (7x + 6):-

→ $7x + 6 = 0$

→ $7x = -6$

→ $\bf{x = \dfrac{-6}{7}}$

$\therefore$Therefore, the 2 zeroes of the given Polynomial are ::

→ $\bf{x = 1}$

→ $\bf{x = \dfrac{-6}{7}}$

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• Verification:-

1.] Compare 7x^2 - x - 6 with ax² + bx + c

→ a = 7

→ b = -1

→ c = -6

2.] Sum of Zeroes =

$1 + \bigg[\dfrac{-6}{7}\bigg]$

→ $\dfrac{7-6}{7}$

→ $\dfrac{1}{7}$

$\implies\bf{\bigg[\dfrac{-1}{7}\bigg]} = \bf{\dfrac{-b}{a}}$

3.] Product of Zeroes =

$1 \times \dfrac{-6}{7}$

→ $\dfrac{-6}{7}$

$\implies\bf{\dfrac{-6}{7}} = \bf{\dfrac{c}{a}}$

Hence, Verified

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