Alph β are the zeros of the polynomial p(x)=3x^2+5x+7 then find the value of 1/alph^2 +1/beta^2

α & β are the zeroes of the Polynomial 3x² + 5x + 7.

The value of 1/α² + 1/β²

On comparing with standard form of a Quadratic equation i.e., ax² + bx + c = 0 ;

Let,

a = 3

b = 5

c = 7

We know that,

Sum of the zeroes = - b/a

⟹ α + β = - 5/3 -- equation (1)

Product of the zeroes = c/a

⟹ αβ = 7/3 -- equation (2)

Now,

We have to find:

⟹ 1/α² + 1/β²

Taking LCM we get,

⟹ (β² + α²)/(α²β²)

we know that,

(a + b)² = a² + b² + 2ab

⟹ (a + b)² - 2ab = a² + b²

⟹ (α + β)² - 2αβ = α² + β²

Again using aⁿ * bⁿ = (ab)ⁿ we get,

⟹ [ (α + β)² - 2αβ ] / (αβ)²

Putting the values from equations (1) & (2) we get,

⟹ [ ( - 5/3)² - 2(7/3) ] / (7/3)²

⟹ (25/9 - 14/3) * (3/7)²

⟹ (25 - 3 * 14)/3 * (9/49)

⟹ ( - 17)/3 * 9/49

⟹ - 51/49

∴ The value of 1/α² + 1/β² is - 51/49.

I hope it will help you.

Regards.

siso your answer refer in pic

ANSWER

α+β=24

α−β=8

(α−β)

2

=(α+β)

2

−4αβ=8

2

or,24

2

−4αβ=64

or,4αβ=576−64=512

or,αβ=128

⇒x

2

−(α+β)x+αβ=0

or,x

2

−24x+128=0

Hence x

2

−24x+128=0 is the quadratic polynomial for the required condition

i dont get the question

step-by-step explanation: