Question

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Answer 1

Answer:

Given polynomial : f(x) = x² - px + q

On comparing with ax² + bx + c, we get

• a = 1, b = - p, c = q

Now,

• Sum of zeroes = α + β = - b/a

→ - (- p)/1

→ p/1

→ p

• Product of zeroes = αβ = c/a

→ q/1

→ q

To Prove :

${\sf{\ \ {\dfrac{\alpha^2}{\beta^2}} + {\dfrac{\beta^2}{\alpha^2}} = {\dfrac{p^4}{q^2}} - {\dfrac{4p^2}{q}} + 2}}$

L.H.S. = ${\sf{ {\dfrac{\alpha^2}{\beta^2}} + {\dfrac{\beta^2}{\alpha^2}}}}$

$\implies{\sf{ {\dfrac{ \alpha^2 (\alpha^2) + \beta^2(\beta^2)}{(\beta^2)(\alpha^2)}}}}$

$\implies{\sf{ {\dfrac{\alpha^4 + \beta^4}{(\alpha \beta)^2}}}}$

• We can write this as :

$\implies{\sf{ {\dfrac{ (\alpha^2)^2 + (\beta^2)^2}{(\alpha \beta)^2}}}}$

• Identity : (a + b)² = a² + b² + 2ab

From this, we get [a² + b² = (a + b)² - 2ab]

Here, a = α², b = β²

$\implies{\sf{ {\dfrac{(\alpha^2 + \beta^2)^2 - 2(\alpha^2)(\beta^2) }{ (\alpha \beta)^2}}}}$

• We can write this as :

$\implies{\sf{ {\dfrac{[(\alpha)^2 + (\beta)^2]^2 - 2(\alpha^2)(\beta^2)}{(\alpha \beta)^2}}}}$

• Identity : (a + b)² = a² + b² + 2ab

From this, we get [a² + b² = (a + b)² - 2ab]

Here, a = α, b = β

$\implies{\sf{ {\dfrac{ [ (\alpha + \beta)^2 - 2 \alpha \beta ]^2 - 2 (\alpha \beta)^2}{ (\alpha \beta)^2}}}}$

• Putting known values.

$\implies{\sf{ {\dfrac{ [(p)^2 - 2(q)]^2 - 2(q)^2}{(q)^2}}}}$

$\implies{\sf{ {\dfrac{[p^2 - 2q]^2 - 2q^2}{q^2}}}}$

• Identity : (a - b)² = a² + b² - 2ab

Here, a = p², b = 2q

$\implies{\sf{ {\dfrac{(p^2)^2 + (2q)^2 - 2(p^2)(2q) - 2q^2}{q^2}}}}$

$\implies{\sf{ {\dfrac{p^4 + 4q^2 - 4p^2q - 2q^2}{q^2}}}}$

• We can write this as :

$\implies{\sf{ {\dfrac{p^4}{q^2}} + {\dfrac{4q^2}{q^2}} - {\dfrac{4p^2q}{q^2}} - {\dfrac{2q^2}{q^2}}}}$

• Cancelling out the common terms.

$\implies{\sf{ {\dfrac{p^4}{q^2}} + 4 - {\dfrac{4p^2}{q}} - 2}}$

$\implies{\sf{ {\dfrac{p^4}{q^2}} - {\dfrac{4p^2}{q}} + 4 - 2}}$

$\implies{\sf{ {\dfrac{p^4}{q^2}} - {\dfrac{4p^2}{q}} + 2}}$

= R.H.S.

Hence, proved !!