Question

Given that α and β are the zeroes of the polynomial p(x) = 3x2 - 6x - 1. Find the polynomial whose zeroes are 1/α² and 1/β²

Answer 1

Step-by-step explanation:

find the roots of first eqn

then,

$1 \div { \alpha }^{2} + 1 \div { \beta }^{2} = ( { \beta }^{2} + { \alpha }^{2} ) \div {( \alpha \beta )}^{2}$

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

sum of the roots are

$( {( \alpha + \beta )}^{2} - 2 \alpha \beta ) \div {( \alpha \beta )}^{2}$

products of the roots are

$1 \div {( \alpha \beta )}^{2}$

put the value of Alpha and beta from the given quadratic equation.

you will get the sum and product of roots of desired quadratic equation.

try it, if you didn't get it message me i will give you the

solution.

Answer 2

Answer:  All polynomials of form a(x² - 42x + 9), where a ≠ 0, will have 1/α² and 1/β² as their zeroes.

Step-by-step explanation:

p(x) = 3x² - 6x - 1 , roots ⇒ α,β

Sum of roots = α + β

$\frac{-(-6)}{3} = \alpha + \beta$

$2 = \alpha + \beta$

Product of roots = αβ

$\frac{-1}{3} = \alpha \beta$

Let the polynomial whose roots are 1/α² and 1/β² be h(x)

h(x) = ax² + bx + c, a ≠ 0, roots ⇒ 1/α², 1/β²

Sum of roots = $\frac{1}{\alpha^{2} } + \frac{1}{\beta^{2} }$

$\frac{-b}{a} = \frac{1}{\alpha^{2} } + \frac{1}{\beta^{2} }$

$\frac{-b}{a} = \frac{1}{\alpha^{2} } + \frac{1}{\beta^{2} }$

$\frac{-b}{a} = \frac{\alpha^{2} +\beta^{2}}{\alpha^{2} \beta^{2} }$

Product of roots = $\frac{1}{\alpha^{2}\beta^{2} }$

$\frac{c}{a} = \frac{1}{\alpha^{2} } \frac{1}{\beta^{2} }$

$\frac{c}{a} = \frac{1}{\alpha^{2} \beta^{2} }$

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

$(2)^{2} = \alpha^{2} +\beta^{2} +2(\frac{-1}{3} )$

$\frac{14}{3} = \alpha^{2} +\beta^{2}$

$\frac{1}{\alpha^{2}\beta^{2} } = (\frac{1}{\alpha \beta } )^{2}$

$\frac{1}{\alpha^{2}\beta^{2} } = (-3 )^{2}$

$\frac{1}{\alpha^{2}\beta^{2} } = 9$

$\frac{-b}{a} = \frac{\alpha^{2} +\beta^{2}}{\alpha^{2} \beta^{2} }$

Therefore,

⇒  $\frac{-b}{a} =\frac{\frac{14}{3} }{\frac{1}{9} }$

⇒  $\frac{-b}{a} = \frac{14}{3}*9$

⇒  $\frac{-b}{a} = 42$

⇒  $b = -42a$

$\frac{c}{a} = \frac{1}{\alpha^{2} \beta^{2} }$

Therefore,

⇒ $\frac{c}{a} = 9$

⇒ $c = 9a$

h(x) = ax² + bx + c

⇒ h(x) = ax² - 42ax + 9a

⇒ h(x) = a(x² - 42x + 9)

Hence, all polynomials of form a(x² - 42x + 9), where a ≠ 0, will have 1/α² and 1/β² as their zeroes.

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