Question

Find the zeroes of quadratic polynomial 8x^2 + 14x + 3 and verify the relationship between the zeroes

and the coefficients.​

Answer 1

Answer:

$\large\boxed{\sf{-\dfrac{1}{4}\;\;\;and\;\;\;-\dfrac{3}{2}}}$

Step-by-step explanation:

Given a quadratic polynomial,

$8 {x}^{2} + 14x + 3$

Here, we have,

• a = 8
• b = 14
• c = 3

To find the zeroes, lets eauate the polynomial to 0.

Therefore, we will get,

$= > 8 {x}^{2} + 14x + 3 = 0$

Now, factorising the above equation, we get,

$= > 8 {x}^{2} + 2x + 12x + 3 = 0 \\ \\ = > 2x(4x + 1) + 3(4x + 1) \\ \\ = > (4x + 1)(2x + 3) = 0 \\ \\ = > 4x + 1 = 0 \: \: \: and \: \: \: 2x + 3 = 0 \\ \\ = > x = - \dfrac{1}{4} \: \: \: \: \: \: and \: \: \: \: \: \: \: x = - \dfrac{3}{2}$

Therefore, the zeroes are -1/4 and -3/2.

Verification:-

We know that,

Sum of zeroes = -b/a = -14/8 = -7/4

=> (-1/4) + (-3/2)= -1/4 - 6/4 = -7/4

And

Product of zeroes = c/a = 3/8

=> (-1/4) × (-3/2) = 3/8

Hence, verified.

Answer 2

Answer:

-1/2 and -3/2 is the correct answer