Question

If alpha and beta are the zeros of the quadratic polynomial f(x)=x^2-px+a,prove that alpha^2/beta^2+beta ^2/alpha ^2=p^4/q^2-4p^2/q+2.

Answer 1

Correct Question:

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

Step-by-step explanation:

Given : f(x) = x² - px + q

On comparing this with ax² + bx + c, we get

• a = 1, b = - p, c = q

Now,

• Sum of zeroes = α + β = - b/a

→ α + β = - (- p)/1

→ α + β = p

• Product of zeroes = αβ = c/a

→ αβ = q/1

→ αβ = q

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To Prove :

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

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

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

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

• We can write this as :

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

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${\boxed{\tt{\bigstar \ \ Identity \ : \ (a + b)^2 = a^2 + b^2 + 2ab}}}$

${\tt{From \ this, \ we \ get \ [ a^2 + b^2 = (a + b)^2 - 2ab}}$

${\tt{Here, \ a = \alpha^2, \ b = \beta^2}}$

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

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${\boxed{\tt{\bigstar \ \ Identity \ : \ (a + b)^2 = a^2 + b^2 + 2ab}}}$

${\tt{From \ this, \ we \ get \ [ a^2 + b^2 = (a + b)^2 - 2ab}}$

${\tt{Here, \ a = \alpha, \ b = \beta}}$

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

• Putting known values.

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

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

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${\boxed{\tt{\bigstar \ \ Identity \ : \ (a - b)^2 = a^2 + b^2 - 2ab}}}$

${\tt{Here, \ a = p^2, \ b = 2q}}$

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

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

• We can write this as :

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

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

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

= R.H.S.

Hence, verified !!