Question

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) .

Answer 1

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

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@ Twilight Astro ✌️☺️

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Answer 2

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