Question

Find the zeroes of the quadratic polynomial 6x
2 + 7x + 2 and verify the
relationship between the zeroes and the coefficients.​

Answer 1

Answer:

6x² + 7x + 2

We know that,

a + b = 7

a × b = 12

3 + 4 = 7

3 × 4 = 12

6x² + 3x + 4x + 2 = 0

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

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

2x + 1 = 0 3x + 2 = 0

2x = -1 3x = -2

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

So,alpha = -1/2 and beta = -2/3

alpha + beta = -1/2 + -2/3

= -3-4/6

= -7/6

-b/a = -7/6

So, alpha + beta = -b/a

alpha × beta = -1/2 × -2/3

= 2/6

c/a = 2/6

So, alpha × beta = c/a

Hence, relationship is verified.

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Answer 2

Answer:

Let P(x) = 6x^2 + 7x + 2

Zeroes of P(x) are given by

P(x) = 0

➳ 6x^2 + 7x + 2 = 0

➳ 6x^2 + 4x + 3x + 2 = 0

➳ 2x(3x + 2) + 1 (3x + 2) = 0

➳ (3x + 2) (2x + 1) = 0

➳ 3x + 2 = 0

Or, 2x + 1 = 0

➬ x = - 2/3 or x = - 1/2

Hence, the zeroes of P(x) are given by

- 2/3 and - 1/2 .

Comparing P(x) with ax^2 + bx + c, we get

a = 6, b = 7, c = 2.

Now, sum of Zeroes

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

= - ( 7/6) = - b/a

Product of Zeroes

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

= 1/3 = 2/6 = c/a

Hence, the relationship between the Zeroes and coefficients is VERIFIED.

Step-by-step explanation:

@Genius