Question

If alpha and beta are the zeroes of 2x2-4x+5.find , alpha3+beta3

Answer 1

$\bf Given:\ \sf \alpha\ and\ \beta\ are\ the\ zeroes\ of\ the\ polynomial\ 2x^2\ -\ 4x\ +\ 5.\\\\\bf To\ find:\ \sf The\ values\ of\ \alpha^3\ +\ \beta^3.\\\\\bf Answer:\\\\\sf Suppose\ ax^2\ +\ bx\ +\ c\ =\ 0\ is\ a\ polynomial.\\\\The\ sum\ of\ the\ zeroes\ of\ that\ polynomial\ =\ \dfrac{-b}{a}.\\\\The\ product\ of\ the\ zeroes\ =\ \dfrac{c}{a}.\\\\\\From\ the\ equation\ we\ have,\ a\ =\ 2,\ b\ =\ -4\ and\ c\ =\ 5.\\\\\\ We\ know\ that\ \alpha\ and\ \beta\ are\ the\ zeroes.$

$\sf \implies\ \alpha\ +\ \beta\ =\ \dfrac{4}{2}\ =\ 2.\\\\\\\implies\ \alpha\ \times\ \beta\ =\ \dfrac{5}{2}.\\\\\\\\Now, \alpha^3\ +\ \beta^3\ =\ (\alpha\ +\ \beta)^3\ -\ 3\alpha\beta(\alpha\ +\ \beta).\\\\\\\bf Using\ the\ values\ we\ have\ in\ this\ formula,\\\\\\\sf \alpha^3\ +\ \beta^3\ =\ (2)^3\ -\ 3\ \times\ \dfrac{5}{2}\ \bigg(2\bigg)\\\\\\\alpha^3\ +\ \beta^3\ =\ 8\ -\ 15\\\\\\\alpha^3\ +\ \beta^3\ =\ -7$

$\sf Therefore,\ \bf \alpha^3\ +\ \beta^3\ =\ -7.$

Answer 2

$\huge{\underline{\underline{\mathfrak{\green{\sf{QUESTION}}}}}}$.

$\sf{\:If \alpha \:and\: \beta \:are\: the \:zeroes \:of \:(2x^2-4x+5)\:.find\: , \alpha^3\:+\beta^3}$

$\huge{\underline{\underline{\mathfrak{\green{\sf{SOLUTION}}}}}}$.

$\red{\sf{\underline{\:Given}}}$

$\star{\sf{\:(2x^2-4x+5\:=0)}}$...(1)

$\star{\sf{\:\alpha\:and\:\beta\:are\:zeros}}$

$\pink{\sf{\underline{\:Find}}}$

$\star{\sf{\:(\alpha^3+\beta^3)}}$

$\huge{\underline{\underline{\mathfrak{\green{\sf{EXPLANaTION}}}}}}$.

We Have,

$\pink{\boxed{\boxed{\sf{\:Sum\:of\:zero\:=\frac{-(Coefficient\:of\:x)}{(Coefficient\:of\:x^2)}}}}}$

$\mapsto\sf{\:(\alpha+\beta)\:=\:\dfrac{-(-4)}{2}}$

$\mapsto\sf{\:(\alpha+\beta)\:=\:2}$....(2)

Again,

$\pink{\boxed{\boxed{\sf{\:Product\:of\:zero\:=\frac{(Constant\:part)}{(Coefficient\:of\:x^2)}}}}}$

$\mapsto\sf{\:(\alpha\beta)\:=\:\dfrac{5}{2}}$.......(3)

____________________

We Have,

$\green{\sf{\:(\alpha^3+\beta^3)\:=\:(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)}}$

$\:\:\:\:\:\red{\sf{\:( Keep\:Value\:by\:equ(1)\:and\:(2)\:)}}$

$\mapsto\sf{\:(\alpha^3+\beta^3)\:=\:(2)^3-3\times\dfrac{5}{\cancel{2}}\times \cancel{2}}$

$\mapsto\sf{\:(\alpha^3+\beta^3)\:=\:8-15}$

$\mapsto\pink{\sf{\:(\alpha^3+\beta^3)\:=\: -7}}$

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