Question

if α & β are roots of 5x^2 - 6x + 3= 0 form a quadratic equation whose roots are α^3β ,αβ^3.​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Quadratic\:eqn\to 625x^{2}-90x+81=0}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies {5x}^{2} - 6x + 3 = 0 \\ \\ \tt: \implies \alpha \: and \: \beta \: are \: zeroes \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Quadratic \: equation \: whose \: zeroes \: { \alpha }^{3} \beta \: and \: \alpha { \beta }^{3} =$

• According to given quetion :

$\tt \circ \: {5x}^{2} - 6x + 3 = 0 \\ \\ \tt \circ \: a = 5 \: \: \: \: \: \: b = - 6 \: \: \: \: \: \: c = 3 \\ \\ \bold{For \: sum \: of \: zeroes :} \\ \tt: \implies \alpha + \beta = \frac{ - b}{a} \\ \\ \tt: \implies \alpha + \beta = \frac{ - ( - 6)}{5} \\ \\ \green{\tt: \implies \alpha + \beta = \frac{6}{5} } \\ \\ \bold{For \: product \: of \: zeroes : } \\ \tt: \implies \alpha \beta = \frac{c}{a} \\ \\ \green{\tt: \implies \alpha \beta = \frac{3}{5} } \\ \\ \bold{For \: Quadratic \: equation : } \\ \tt: \implies {x}^{2} - (sum \: of \: new \: zeroes)x + (product \: of \: new \: zeroes) \\ \\ \tt: \implies {x}^{2} - ( { \alpha }^{3} \beta + { \alpha \beta }^{3} )x + ( { \alpha }^{3} \beta \times { \alpha \beta }^{3} ) = 0$

$\tt: \implies {x}^{2} - ( \alpha \beta ( { \alpha }^{2} + { \beta }^{2} ))x + { (\alpha \beta) }^{4} \\ \\ \tt: \implies {x}^{2} - ( \alpha \beta (( { \alpha + \beta )}^{2} - 2 \alpha \beta ))x + ( { \alpha \beta) }^{4} = 0 \\ \\ \tt: \implies {x}^{2} - ( \frac{3}{5} ( ( \frac{6}{5} )^{2} - 2 \times \frac{3}{5} ))x + { (\frac{3}{5} )}^{4} = 0 \\ \\ \tt: \implies {x}^{2} - ( \frac{3}{5} ( \frac{36}{25} - \frac{6}{5} ))x + \frac{81}{625} = 0 \\ \\ \tt: \implies {x}^{2} - ( \frac{3}{5} ( \frac{36 - 30}{25} ))x + \frac{81}{625} = 0 \\ \\ \tt: \implies {x}^{2} - ( \frac{3}{5} \times \frac{6}{25} )x + \frac{81}{625} = 0 \\ \\ \tt: \implies {x}^{2} - (\frac{18}{125} )x + \frac{81}{625} = 0 \\ \\ \green{\tt: \implies {625x}^{2} - 90x +81 = 0}$

Answer 2

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{Quadratic \ equation \ with \ required \ roots \ is}$

$\bold{625x^{2}-90x+81=0.}$

$\bold\orange{Given:}$

$\bold{The \ given \ quadratic \ equation \ is}$

$\bold{=>5x^{2}-6x+3=0}$

$\bold{=>\alpha \ and \beta \ are \ roots \ of \ the}$

$\bold{equation.}$

$\bold\pink{To \ find:}$

$\bold{New \ quadratic \ equation \ with \ roots}$

$\bold{=>\alpha^{3}\beta \ and \alpha\beta^{3}}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{The \ given \ quadratic \ equation \ is}$

$\bold{=>5x^{2}-6x+3=0}$

$\bold{Here,a=5, \ b=-6, \ and \ c=3}$

________________________________

$\bold{Sum \ of \ roots=\frac{-b}{a}}$

$\bold{Product \ of \ roots=\frac{c}{a}}$

$\bold{\tt{\therefore{\alpha+\beta=\frac{6}{5}...(1)}}}$

$\bold{\tt{\therefore{\alpha\beta=\frac{3}{5}...(2)}}}$

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

$\bold{\alpha^{2}+\beta^{2}=(\frac{6}{5})^{2}-2(\frac{3}{5})}$

$\bold{...from \ (1) \ and \ (2)}$

$\bold{\alpha^{2}+\beta^{2}=\frac{36}{25}-\frac{6}{5}}$

$\bold{\tt{\therefore{\alpha^{2}+\beta^{2}=\frac{36-30}{25}}}}$

$\bold{\tt{\therefore{\alpha^{2}+\beta^{2}=\frac{6}{25}...(3)}}}$

_____________________________________

$\bold{Roots \ of \ new \ equation \ are}$

$\bold{\alpha^{3}\beta \ and \alpha\beta^{3}}$

$\bold{Let \ M \ and \ N \ be \ roots \ of \ new \ equation}$

$\bold{M+N=\alpha^{3}\beta+\alpha\beta^{3}}$

$\bold{\tt{\therefore{M+N=\alpha\beta(\alpha^{2}+\beta^{2})}}}$

$\bold{...from \ (3) \ and \ (2)}$

$\bold{M+N=\frac{3}{5}×\frac{6}{25}}$

$\bold{\tt{\therefore{M+N=\frac{18}{125}...(4)}}}$

$\bold{M×N=\alpha^{3}\beta×\alpha\beta^{3}}$

$\bold{\tt{\therefore{M×N=(\alpha\beta)^{4}}}}$

$\bold{M×N=(\frac{3}{5})^{4}}$

$\bold{\tt{\therefore{M×N=\frac{81}{625}...(5)}}}$

___________________________________

$\bold{Quadratic \ equation \ can \ be \ written \ as}$

$\bold{=>x^{2}-(Sum \ of \ roots)x+(Product \ of \ roots)=0}$

$\bold{=>x^{2}-(M+N)x+(M×N)=0}$

$\bold{From \ (4) \ and \ (5)}$

$\bold{=>x^{2}-\frac{18x}{125}+\frac{81}{625}=0}$

$\bold{Multiply \ equation \ by \ 625 \ throughout}$

$\bold{we \ get,}$

$\bold{=>625x^{2}-90x+81=0}$

$\bold\purple{\tt{\therefore{Quadratic \ equation \ with \ required \ roots \ is}}}$

$\bold\purple{625x^{2}-90x+81=0.}$