Question

solve:if
$\alpha$
and
$\beta$
are the zeroes of the quadratic polynomial
$p(y) = 3 {y}^{2} - 6y + 4$
find the value of
$\alpha/ \beta + \beta / \alpha$

​

Answer 1

Answer:

The answer is in the attachment,

Answer 2

Step-by-step explanation:

Given -

α and β are zeroes of polynomial

p(y) = 3y² - 6y + 4

To Find -

Value of α/β + β/α

3y² - 6y + 4

here,

a = 3

b = -6

c = 4

As we know that :-

• α + β = -b/a

» -(-6)/3

» 6/3

• » 2

And

• αβ = c/a

» 4/3

Now,

α + β = 2

Squaring both sides :-

(α + β)² = (2)²

» α² + 2αβ + β² = 4

» α² + β² + 2(4/3) = 4

» α² + β² + 8/3 = 4

» α² + β² = 4 - 8/3

» α² + β² = 12 - 8/3

• » α² + β² = 4/3

Now,

α/β + β/α

» α² + β²/αβ

» 4/3 ÷ 4/3

• » 1

Hence,

The value of α/β + β/α is 1