Math, asked by thomasvikrant, 10 months ago

If alpha and beta are the zeroes of the quadratic polynomial f(x)=6x²+x-2,find the value of alpha³+beta³

Answers

Answered by Anonymous
19

\sf\red{\underline{\underline{Answer:-}}}

\sf{The \ value \ of \ \alpha^{3}+\beta^{3} \ is \ \frac{-37}{216}}

\sf\orange{Given:}

\sf{The \ given \ polynomial \ is}

\sf{\implies{f(x)=6x^{2}+x-2}}

\sf{\implies{Zeroes \ are \ \alpha \ and \ \beta}}

\sf\pink{To \ find:}

\sf{The \ value \ of \ \alpha^{3}+\beta^{3}}

\sf\green{\underline{\underline{Solution:}}}

\sf{The \ given \ polynomial \ is}

\sf{\implies{f(x)=6x^{2}+x-2}}

\sf{Here, \ a=6, \ b=1 \ and \ c=-2}

\sf{Sum \ of \ zeroes=\frac{-b}{a}}

\sf{\therefore{\alpha+\beta=\frac{-1}{6}...(1)}}

\sf{Product \ of \ zeroes=\frac{c}{a}}

\sf{\therefore{\alpha\beta=\frac{-2}{6}}}

\sf{\therefore{\alpha\beta=\frac{-1}{3}...(2)}}

\sf{According \ to \ identity}

\sf{a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)}

\sf{\therefore{\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3(\alpha\beta)(\alpha+\beta)}}

\sf{....from \ (1) \ and \ (2)}

\sf{\therefore{\alpha^{3}+\beta^{3}=(\frac{-1}{6})^{3}-3(\frac{-1}{3})(\frac{-1}{6})}}

\sf{\therefore{\alpha^{3}+\beta^{3}=(\frac{-1}{6})^{3}-(\frac{1}{6})}}

\sf{\therefore{\alpha^{3}+\beta^{3}=\frac{-1}{216}-\frac{1}{6}}}

\sf{\therefore{\alpha^{3}+\beta^{3}=\frac{-1-36}{216}}}

\sf{\therefore{\alpha^{3}+\beta^{3}=\frac{-37}{216}}}

\sf\purple{\tt{\therefore{The \ value \ of \ \alpha^{3}+\beta^{3} \ is \ \frac{-37}{216}}}}

Answered by Anonymous
19

 \large\bf\underline \orange{Given:-}

  • p(x) = 6x² + x -2

 \large\bf\underline \orange{To \: find:-}

  • value of α³ + β³

 \huge\bf\underline \green{Solution:-}

★ Given equation = 6x² + x -2

  • a = 6
  • b = 1
  • c = -2

α and β are the roots of the given polynomial.

sum of zeroes = coefficient of x/coefficient of x²

  • α + β = -b/a

⠀⠀⠀⠀⠀➝ α + β = -1/6

  • α β = c/a

⠀⠀⠀⠀⠀➝ αβ = -2/6

⠀⠀⠀⠀⠀➝ α β = -1/3

 \pink{\boxed{  \bf({a}+  {b}) {}^{3}  =  {a}^{3}  + b {}^{3}   + 3ab (a + b)}} \\  \\  \longmapsto \rm \: ({a}+  {b}) {}^{3}  =  {a}^{3}  + b {}^{3}   + 3ab (a + b) \\  \\ \longmapsto \rm \: {a}^{3}  + b {}^{3}  = (a + b) {}^{3}  - 3ab(a + b) \\  \\ \longmapsto \rm \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \: =  ( \alpha  +  \beta ) {}^{3}  - 3 \alpha  \beta ( \alpha  +  \beta ) \\  \\ \longmapsto \rm \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \: =( \frac{ - 1}{6} ) {}^{3}  - \cancel 3 \times  \frac{ - 1}{ \cancel3} ( \frac{ - 1}{6} ) \\  \\ \longmapsto \rm \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \: = \frac{ - 1}{216}  -  \frac{1}{6}  \\  \\ \longmapsto \rm \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \: = \frac{ - 1 - 36}{216}  \\  \\ \longmapsto \bf \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \: = \frac{ - 37}{216}

So,

α³ + β³ = -37/216

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