Question

if alpha and beta are the zeroes of the polynomial x^+4x+3,find the polynomial whose zeroes are 1+beta/alpha and 1+alpha/bete​

Answer 1

α and β are the two zeroes of the polynomial x² + 4x + 3.

So,

By using sum of zeroes and product of zeroes

α + β = -4 and αβ = 3

★ Sum of zeroes:

$1 + \frac{ \beta }{ \alpha } + 1 + \frac{ \alpha }{ \beta }$

$\rightarrow \frac{ \alpha \beta + \beta {}^{2} + \alpha \beta + { \alpha }^{2} }{ \alpha \beta }$

$\rightarrow \frac{ { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta }{ \alpha \beta }$

$\rightarrow \: \frac{( \alpha + \beta ) {}^{2} }{ \alpha \beta }$

$\rightarrow \: \frac{( - 4) {}^{2} }{3}$

$\rightarrow \: \frac{16}{3}$

★ Product of the zeroes:

$(1 + \frac{ \beta }{ \alpha } )(1 + \frac{ \alpha }{ \beta } )$

$\rightarrow \: 1 + \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } + \frac{ \alpha \beta }{ \alpha \beta }$

$\rightarrow \: \frac{2 \alpha \beta + \alpha {}^{2} + { \beta }^{2} }{ \alpha \beta }$

$\rightarrow \: \frac{( \alpha + \beta ) {}^{2} }{ \alpha \beta }$

$\rightarrow \: \frac{ ( - 4) {}^{2} }{3}$

$\rightarrow \: \frac{16}{3}$

But required polynomial:

k {x² - ( sum of zeroes ) x + Product the zeroes}

⇒ x² - 16/3x + 16/3

If k = 3

⇒ 3 { x² - 16/3x + 16/3 }

⇒ 3x2 - 16x + 16

Answer 2

$\huge\mathfrak{Amswer:}$

Given:

• We have been given that α and β are the two zeroes of the polynomial x² + 4x + 3.

To Find:

• We need to find the polynomial whose zeroes are 1 + β/α and 1 + α/β.

Solution:

The given polynomial is:

p(x) = x² + 4x + 3.

a = 1, b = 4 and c = 3

Sum of zeroes (α + β)

= -b/a

= -4/1______(1)

Product of zeroes (αβ)

=c/a

= 3/1______(2)

Now, using the sum and product of zeroes, we have

Sum of zeroes:

$\sf{ = 1 + \dfrac{ \beta }{ \alpha } + 1 + \dfrac{ \alpha }{ \beta } }$

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

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

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

Substituting the values from equation 1 and 2, we have

$\mapsto\sf{ \dfrac{ {( - 4)}^{2} }{3} }$

$\longrightarrow \sf{ \dfrac{16}{3} }$

Product of zeroes:

$\sf{(1 + \dfrac{ \beta }{ \alpha})(1 + \dfrac{ \alpha }{ \beta }) }$

$\implies\sf{1 + \dfrac{ \alpha }{ \beta } + \dfrac{ \beta }{ \alpha } + \dfrac{ \alpha \beta }{ \alpha \beta } }$

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

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

$\implies\sf{ \dfrac{ {( - 4)}^{2} }{3} }$

$\longrightarrow\sf{ \dfrac{16}{3} }$

Now, we can find the polynomial by this formula:

$\mapsto\sf{k{ \times {x}^{2} - ( \alpha + \beta )x + \alpha \beta} }$

$\longrightarrow\sf{ {x}^{2} - \dfrac{16}{3} x + \dfrac{16}{3} }$

Putting k = 3, we have

$\longrightarrow\sf{3 \times {x}^{2} - \dfrac{16}{3} x + \dfrac{16}{3} }$

$\longrightarrow\sf{3 {x}^{2} - 16x + 16 }$

Hence, the required polynomial is 3x² - 16x + 16.