Question

If $\alpha and \beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}+px+q$,prove that $\frac{\alpha^{2}}{\beta^{2}} + \frac {\beta^{2}} {\alpha^{2}} = \frac{p^{2}}{q^{2}} - \frac{4p^{2}}{q} +2$

Answer 1

SOLUTION :

Given :  α and β are the roots of the quadratic polynomial  f(x)= x²  - px + q

On comparing with ax² + bx + c,

a = 1 , b = -p , c = q

Sum of the zeroes = −coefficient of x / coefficient of x²

α + β  = -b/a = -(-p)/1 = p

α+β = p ……………………….(1)  

Product of the zeroes = constant term/ Coefficient of x²

αβ = c/a = q/1 = q

α×β = q ………………..(2)

LHS = α²/β² + β²/α²

= α⁴ +β⁴ / α²β²     [By taking L.C.M]

As we know that,  a⁴ +b⁴ =  (a² + b²)² – 2a²b²

= (α² + β²)² – 2(αβ)² /(αβ)²

= [(α+β)²–2αβ]² –2 (αβ)² /(αβ)²

[ a² + b² =  (a + b)² - 2ab  ]

= [p² –2q]² –2(q)²/(q)²

= [(p²)² + (2q)² - 2× p²× 2q] - 2q² / q²

= (p⁴ + 4q² - 4p²q)–2q² /q²

[  (a -  b)² =  a² + b²  - 2ab  ]

= (p⁴ + 4q² –2q² - 4p²q)  /q²

= (p⁴ + 2q² - 4p²q)  /q²

= p⁴/q² +2q²/q² - 4p²q/q²

α²/β² + β²/α²  = p⁴/q²  - 4p²/q + 2

LHS = RHS

Hence, proved.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Answer:

SOLUTION :

Given :  α and β are the roots of the quadratic polynomial  f(x)= x²  - px + q

On comparing with ax² + bx + c,

a = 1 , b = -p , c = q

Sum of the zeroes = −coefficient of x / coefficient of x²

α + β  = -b/a = -(-p)/1 = p

α+β = p ……………………….(1)  

Product of the zeroes = constant term/ Coefficient of x²

αβ = c/a = q/1 = q

α×β = q ………………..(2)

LHS = α²/β² + β²/α²

= α⁴ +β⁴ / α²β²     [By taking L.C.M]

As we know that,  a⁴ +b⁴ =  (a² + b²)² – 2a²b²

= (α² + β²)² – 2(αβ)² /(αβ)²

= [(α+β)²–2αβ]² –2 (αβ)² /(αβ)²

[ a² + b² =  (a + b)² - 2ab  ]

= [p² –2q]² –2(q)²/(q)²

= [(p²)² + (2q)² - 2× p²× 2q] - 2q² / q²

= (p⁴ + 4q² - 4p²q)–2q² /q²

[  (a -  b)² =  a² + b²  - 2ab  ]

= (p⁴ + 4q² –2q² - 4p²q)  /q²

= (p⁴ + 2q² - 4p²q)  /q²

= p⁴/q² +2q²/q² - 4p²q/q²

α²/β² + β²/α²  = p⁴/q²  - 4p²/q + 2

LHS = RHS

Hence, proved.

HOPE THIS ANSWER WILL HELP YOU….