Question

if alpha and beta are the zeros of the polynomial ax square + bx + c .
then find the value of
1)alpha/beta + beta/alpha
2)1/alpha cube + 1/beta cube​

Answer 1

Solution:

→ α + ß = - Coefficient of x/Coefficient of x²

→ αß = Constant Term/Coefficient of x²

In ax² + bx + c we have

→ α + ß = - b/a

→ αß = c/a

1) α/ß + ß/α

→ (α² + ß²)/αß → [(α + ß)² - 2αß]/αß

→ [(- b/a)² - 2c/a]/(c/a)

→ (b²/a² - 2ac/a²)/(c/a)

→ (b² - 2ac)/a² × a/c

→ (b² - 2ac)/ac or b²/ac - 2

2) 1/α³ + 1/ß³

→ (α³ + ß³)/(αß)³

→ (α + ß)(α² + ß² - αß)/(αß)³

→ (- b/a){(α + ß)² - 2αß - αß}/(αß)³

→ (- b/a){b²/a² - 3ac/a²}/(c/a)

→ - b/a × (b² - 3ac)/a² × a/c

→ - b/a × (b² - 3ac)/ac

→ - b(b² - 3ac)/a²c

Answer 2

Answer:

$\large\boxed{\sf{(1)\frac{ {b}^{2} - 2ac}{ac} }}$

$\large\boxed{\sf{(2)\frac{3abc - {b}^{3} }{ {c}^{3} } }}$

Step-by-step explanation:

Given Polynomial:

$a {x}^{2} + bx + c$

Also, the zeroes are alpha and beta.

We know that:

$\alpha + \beta = - \frac{b}{a}$

and

$\alpha \beta = \frac{c}{a}$

Now, coming to the Question

$(1) \dfrac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } \\ \\ = \dfrac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } \\ \\ = \dfrac{ {( \alpha + \beta )}^{2} - 2 \alpha \beta }{ \alpha \beta } \\ \\ = \dfrac{ {( - \frac{b}{a} )}^{2} - 2 \frac{c}{a} }{ \frac{c}{a} } \\ \\ = \dfrac{ \frac{ {b}^{2} }{ {a}^{2} } - \frac{2ac}{ {a}^{2} } }{ \frac{ac}{ {a}^{2} } } \\ \\ = \dfrac{ {b}^{2} - 2ac}{ac}$

$(2) \frac{1}{ { \alpha }^{3} } + \frac{1}{ { \beta }^{ 3} } \\ \\ = \dfrac{ { \alpha }^{3} + { \beta }^{3} }{ {( \alpha \beta )}^{3} } \\ \\ = \dfrac{( \alpha + \beta )( { \alpha }^{2} - \alpha \beta + { \beta }^{2} )}{ {( \alpha \beta )}^{3} } \\ \\ = \dfrac{ - \frac{b}{a}( {( \alpha + \beta ) }^{2} - 3 \alpha \beta) }{ {( \frac{c}{a}) }^{3} } \\ \\ = \dfrac{ - \frac{b}{a} ( \frac{ {b}^{2} }{ {a}^{2} } - 3 \frac{c}{a} )}{ \frac{ {c}^{3} }{ {a}^{3} } } \\ \\ = \dfrac{ - \frac{ {b}^{3} }{ {a}^{3} } + \frac{3abc}{ {a}^{3} } }{ \frac{ {c}^{3} }{ {a}^{3} } } \\ \\ = \dfrac{3abc - {b}^{3} }{ {c}^{3} }$