Question

If α and β are the zeros of the quadratic polynomial f(x) = x² - px + q, prove that (α²/β²) + (β²/α²) = (p⁴/q²) - (4p²/q) + 2

Answer 1

(α²/β²) + (β²/α²)  =  p⁴/q²  - 4p²/q  + 2 if α and β are the zeros of the quadratic polynomial f(x) = x² - px + q,

Step-by-step explanation:

α and β are the zeros of the quadratic polynomial

f(x) = x² - px + q

α +  β  = p

αβ  = q

LHS = (α²/β²) + (β²/α²)

= (α⁴  + β⁴)/(α²β²)

= ((α² + β²)² - 2α²β² )/(αβ)²

= ( ((α +  β)² - 2αβ)² - 2(αβ)² )/(αβ)²

= ( (p²  - 2q)² - 2q²)/q²

= ( p⁴ + 4q²  - 4p²q - 2q²)/q²

= ( p⁴  - 4p²q + 2q²)/q²

= p⁴/q²  - 4p²/q  + 2

= RHS

QED

Proved

(α²/β²) + (β²/α²)  =  p⁴/q²  - 4p²/q  + 2

Learn more:

If one root of the quadratic equation 2x^2+kx-6=0 is 2 then fin find ...

https://brainly.in/question/7564820

If one root of the quadratic equation kx^2-3x-1=0 is 1/2 the find value ...

https://brainly.in/question/7551719

Find the roots of the quadratic equation 2x^2-22x+1=0 - Brainly.in

https://brainly.in/question/8643624

Answer 2

Answer:

Step-by-step explanation:

α and β are the zeros of the quadratic polynomial

f(x) = x² - px + q

α +  β  = p

αβ  = q

LHS = (α²/β²) + (β²/α²)

= (α⁴  + β⁴)/(α²β²)

= ((α² + β²)² - 2α²β² )/(αβ)²

= ( ((α +  β)² - 2αβ)² - 2(αβ)² )/(αβ)²

= ( (p²  - 2q)² - 2q²)/q²

= ( p⁴ + 4q²  - 4p²q - 2q²)/q²

= ( p⁴  - 4p²q + 2q²)/q²

= p⁴/q²  - 4p²/q  + 2

LHS= RHS

Hence Proved

(α²/β²) + (β²/α²)  =  p⁴/q²  - 4p²/q  + 2