Question

$\large \purple \bigstar \: \color{red} \boxed{ \sf \color{maroon}{Quality \: Question : }}$
If α and β be two zeros of the quadratic polynomial ax² + bx + c, then evaluate :

$\sf(i) \: \alpha^{2} + \beta^{2} \\ \\ \sf(ii) \: \alpha^{3} + \beta^{3} \\ \\ \sf(iii) \: \frac{1}{ \alpha^{3} } + \frac{1}{ \beta^{3} } \\ \\ \sf(iv) \frac{ { \alpha}^{2} }{ \beta } + \frac{ { \beta}^{2} }{ \alpha}$
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Answer 1

Answer:-

Given:

α and β are the roots of ax² + bx + c.

We know that;

Sum of the roots = - b/a

So,

⟹ α + β = - b/a -- equation (1)

Product of the roots = c/a

⟹ αβ = c/a -- equation (2)

Now,

We have to find:

1) α² + β²

We know that;

a² + b² = (a + b)² - 2ab

Hence,

⟹ α² + β² = (α + β)² - 2αβ

⟹ α² + β² = ( - b/a)² - 2(c/a)

[ ∵ From equations (1) , (2). ]

⟹ α² + β² = b²/a² - 2c/a

⟹ α² + β² = (b² - 2ac)/a²

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2) α³ + β³

We know that;

a³ + b³ = (a + b)³ - 3ab(a + b)

⟹ α³ + β³ = (α + β)³ - 3αβ(α + β)

⟹ α³ + β³ = ( - b/a)³ - 3(c/a)(- b/a)

[ ∵ From equations (1) , (2) ]

⟹ α³ + β³ = - b³/a³ + 3bc/a²

⟹ α³ + β³ = (- b³ + 3abc) / a³

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3) (1/α³) + (1/β³)

Taking LCM we get,

⟹ (β³ + α³) / (α³β³)

Substitute the value of α³ + β³ .

⟹ { ( - b³ + 3abc)/a³ } × 1/(αβ)³

⟹ { ( - b³ + 3abc)/a³ } × 1/(c/a)³

⟹ { ( - b³ + 3abc) / a³ } × a³/c³

⟹ ( - b³ + 3abc) / c³

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4) (α²/β) + (β²/α)

Taking LCM we get,

⟹ (α³ + β³) / αβ

Substitute the value of α³ + β³ here.

⟹ { ( - b³ + 3abc) / a³ } × (1/αβ)

⟹ { (- b³ + 3abc) /a³ } × 1/(c/a)

⟹ { (- b³ + 3abc) / a³ } × a/c

⟹ ( - b³ + 3abc) / a²c

Answer 2

Answer:

Required Answer :-

At first we need to know some basics

Like

Sum of zeroes is given by

$\sf \: \alpha + \beta = \dfrac{ - b}{a}$

Product of zeroes is given by

$\sf \: \alpha \beta = \dfrac{ c}{a}$

Now solving them

I]

$\sf \: ( \alpha + \beta {)}^{2} = { \alpha }^{2} - 2 \alpha \beta + \beta {}^{2}$

$\sf \: ( \alpha + \beta {)}^{2} = ( { \alpha + \beta) }^{2} - 2 \alpha \beta$

$\sf \implies \dfrac{ - b}{a}^{2} - 2 \bigg( \dfrac{c}{a} \bigg)$

Since

- × - = +

$\sf \implies \: \dfrac{ {b}^{2} }{a {}^{2} } - \dfrac{2c}{a}$

$\sf \implies \: \dfrac{ {b }^{2} - 2ca }{ {a}^{2} }$

2]

$\sf \: { (\alpha^3 + \beta )}^{3} =(\alpha+ \beta)^3 - 3\alpha\beta (\alpha + \beta)$

By putting value

$\sf \implies \dfrac{-b}{a}^3 - 3\bigg(\dfrac{c}{a}\bigg) \times \dfrac{-b}{a}$

$\sf \implies \: \dfrac{ - {b}^{3} }{ { a}^{3} } - 3 \times \dfrac{c}{a} \times \dfrac{ - b}{a}$

= -b³ + 3abc/a²

3]

$\sf \: \dfrac{ { \alpha }^{3} + { \beta }^{3} }{ { \alpha }^{3} { \beta }^{3} }$

According to the question

$\sf \dfrac{ - {b}^{3} + 3ab}{ { \alpha }^{3}} \times \dfrac{1} {{\dfrac{c}{a} }^{3} }$

$\sf \: \dfrac{ - b + 3abc}{a {}^{3} } \times \dfrac{1}{ {c}^{3} } \times {a}^{3}$

$\sf \: \dfrac{ - b + 3abc}{a {}^{3} } \times \dfrac{ {a}^{3} }{ {b}^{3} }$

$\sf \dfrac{ - b + 3abc}{ {b}^{3}}$

4]

$\sf \dfrac{ - {b}^{3} + 3abc}{ {a}^{3}} \times \dfrac{1}{ \dfrac{a}{c} }$

$\sf \: \dfrac{ - {b}^{3} + 3abc}{ {a}^{3} } \times \dfrac{1}{c} \times a$

$\sf \: \dfrac{ { - b}^{3} + 3abc }{ {a}^{3} } \times \dfrac{a}{c}$

$\sf \dfrac{ - b {}^{3} + 3abc}{ {a}^{2} c}$