Math, asked by shubhambhangdiiya202, 10 months ago

If ∝and β are the zeroes of the quadratric polynomial f(x) =x2 -3x -2,
find a quadratic polynomial whose zeroes are 1/(2α+ β) and 1/(2β+ α) .

Answers

Answered by saivivek16
6

Step-by-step explanation:

Aloha !

 \text { This is Twilight Astro}

Given,.

p(x)=x²-3x-2

Let,

x²-3x-2=0

x²-2x-x-2=0

-x(-x-2)+1(-x-2)=0

(-x-2)(-x+1)=0

-x-2=0 (or) -x+1=0

-x=2. (or) -x= -1

x= -2 (or) x=1

Sum of roots= -2+1= -1

product of roots= -2×1= -2

Quadratic equation:-

X²- (sum of roots) X + product of roots=0

X²-(-1)X+(-2)=0

X²+X-2=0

Thank you

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Answered by Anonymous
5

Aɴꜱᴡᴇʀ

 \huge \sf{}2 0{x}^{2}  - 9x + 1

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Gɪᴠᴇɴ

 \sf{} \alpha  \: and \:  \beta  \: are \: the \: zeros \: of \: the \: polynomial \:   {x}^{2} {- 3x  + 2}

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ᴛᴏ ꜰɪɴᴅ ᴛᴏ ᴘʀᴏᴠᴇ

 \sf{}a \: quadratic \: polynomial \: whose \: zeros \: are \:  \frac{1}{2 \alpha  +  \beta }  \: and \:  \frac{1}{2 \beta  +  \alpha }

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Sᴛᴇᴘꜱ

 \sf{}p(x) =  {x}^{2}  - 3x   + 2 \\ \\   \bf {\sf{}}now \: refer \: to \: the \: attachment \\ \\ \sf{}on \: comparing \: with \: a {x}^{2}  + bx + c = 0 \\  \\ \huge{ \fbox{ \fbox{ \red{ \sf{20 {x}^{2} - 9x + 1 }}}}}

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\huge{\mathfrak{\purple{hope\; it \;helps}}}

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