If-and-are-the-roots-of-the-equation-3x-2-x-2-0-find-the-equation-whose-roots-are-1-2-and-1-2-and-show-that-27-4-11-10-

Question Number 138766 by nadovic last updated on 18/Apr/21

If α and β are the roots of the equation

3 x^{2} + x + 2 = 0, find the equation whose

roots are \frac{1}{α^{2}} and \frac{1}{β^{2}} and show that

27 α^{4} = 11 α + 10 .

Commented by MJS_new last updated on 18/Apr/21

x^{2} + p x + q = 0 with roots α, β

\Rightarrow

x^{2} + \frac{2 q - p^{2}}{q^{2}} x + \frac{1}{q^{2}} = 0 with roots \frac{1}{α^{2}}, \frac{1}{β^{2}}

Answered by bramlexs22 last updated on 18/Apr/21

3 x^{2} + x + 2 = 0 \to {\begin{cases} α \\ β \end{cases}

\Rightarrow 3 {(\sqrt{x})}^{2} + \sqrt{x} + 2 = 0

\Rightarrow \sqrt{x} = - 2 - 3 x

\Rightarrow x = 4 + 12 x + 9 x^{2}

\Rightarrow 9 x^{2} + 11 x + 4 = 0 \to {\begin{cases} α^{2} \\ β^{2} \end{cases}

a n d 4 x^{2} + 11 x + 9 = 0 \to {\begin{cases} \frac{1}{α^{2}} \\ \frac{1}{β^{2}} \end{cases}

Commented by Rasheed.Sindhi last updated on 18/Apr/21

\Rightarrow x = 4 + 12 x + 3 x^{2}

\overset{?}{\Rightarrow} 3 x^{2} + 11 x + 4 = 0 \to {\begin{cases} α^{2} \\ β^{2} \end{cases}

Commented by bramlexs22 last updated on 18/Apr/21

t y p o

Answered by Rasheed.Sindhi last updated on 18/Apr/21

α + β = - 1 / 3, α β = 2 / 3

\frac{1}{α^{2}} + \frac{1}{β^{2}} = \frac{α^{2} + β^{2}}{α^{2} β^{2}} = \frac{{(α + β)}^{2} - 2 α β}{{(α β)}^{2}}

= \frac{{(- 1 / 3)}^{2} - 2 (2 / 3)}{{(2 / 3)}^{2}} = \frac{\frac{1}{9} - \frac{4}{3}}{\frac{4}{9}}

= \frac{1 - 12}{4} = - \frac{11}{4}

\frac{1}{α^{2}} . \frac{1}{β^{2}} = \frac{1}{{(α β)}^{2}} = \frac{1}{{(2 / 3)}^{2}} = \frac{9}{4}

x^{2} - (\frac{1}{α^{2}} + \frac{1}{β^{2}}) x + \frac{1}{α^{2}} . \frac{1}{β^{2}} = 0

x^{2} - (- \frac{11}{4}) x + \frac{9}{4} = 0

4 x^{2} + 11 x + 9 = 0

Answered by physicstutes last updated on 18/Apr/21

α + β = - \frac{b}{a} = - \frac{1}{3}

α β = \frac{c}{a} = \frac{2}{3}

sum of roots : \frac{1}{α^{2}} + \frac{1}{β^{2}} = \frac{{(α + β)}^{2} - 2 α β}{{(α β)}^{2}} = \frac{(\frac{1}{9}) - \frac{4}{3}}{\frac{4}{9}} = \frac{- 11}{4}

product of roots : (\frac{1}{α^{2}}) (\frac{1}{β^{2}}) = \frac{1}{{(α β)}^{2}} = \frac{9}{4}

new equation : x^{2} + \frac{11}{4} x + \frac{9}{4} = 0 \Rightarrow 4 x^{2} + 11 x + 9 = 0