# if alfa and beta are the zeroes of of quadratic polynomial f(x)=x2-3x-2'find quadratic polynomial whose zeroes are 1/2alfa+beta and 1/2beta+alfa

Thx Domi..

• -17
someone has proved this wrong i think...
9/-10+2[9-2*-2]=9/-10+26=9/16

• -3
someone has proved this wrong i think...
sum= 9/-10+2[9-2*-2]=9/-10+26=9/16

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Can someone please elaborate the steps and show me!! :)
• -7
thnx domi

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Thanks domi
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dont know

• -6
What a math domi...wrong calculation
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It is given that a and b are zeros of polynomial f(x)=x2-3x-2. Therefore, a+b=3 ab=-2 Now, the zeroes of the required quadratic polynomial are  Sum of roots = Product of roots =   Now, the required quadratic polynomial is given by: x2 - (Sum of roots)x + (Product of roots) Now, substitute the values
• 1
9/6 and root5 is the correct answer
• -6
bilkul sahi kha shreya saw

• -6
2x2-9x+7
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Itne mushkil questions aane lage to mein to chala gaon pdayi pduyi chhod kr
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Hope this helps

• 21
Domi ur answer is wrong . Plz check my Answer it is 100% correct ...ive checked it also with my tr.
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• -2
100%wrong
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Uday

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If there are any doubts then ask Hope it helps:)

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Domo Sheena both are wrong
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Hi
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Note :

Here I am using m and n instead of Alfa and beta.

Given:

If m and n are the zeroes of

To find :

zeroes are 1/(2m+n) and 1/(2n+m) .

Explanation:

Compare f(x)= -3x-2 with

ax²+bx+c, we get

a = 1 , b = -3 , c = -2

i) m+n = -b/a = -(-3)/1 = 3--(1)

ii) mn = c/a = -2/1 = -2--(2)

iii) + = (m+n)²-2mn

= 3² - 2×(-2)

= 9 + 4

= 13 ----(3)

Now ,

1/(2m+n) , 1/(2n+m) are zeroes

of a polynomial.

1) Sum of the zeroes

= 1/(2m+n) + 1/(2n+m)

= [2n+m+2m+n]/[(2m+n)(2n+m)]

= [3m+3n]/[4mn+2m²+2n²+mn]

= [3(m+n)]/[2(+)+5mn]

= [3×3]/[2×13 + 5(-2)]

= 9/(26-10)

= 9/16 ----(4)

2) Product of the zeroes

= [1/(2m+n) × 1/(2n+m)]

= 1/( 4mn+2m²+2n²+mn)

= 1/[2(+) + 5mn ]

= 1/[ 2×13 + 5(-2)]

= 1/(26-10)

= 1/16 ---(5)

______________________

is

k[-(sum of the zeroes)x+product of the zeroes]

______________________

Here ,

Required polynomial is

k[ -(9/16)x+1/16]

For all real values of k it is true.

let k = 16,

16[-(9/16)x+1/16]

= 16x²-9x+1

Therefore,

Polynomial whose zeroes are

1/(2m+n) and 1/(2n+m) is

16x²-9x+1

• 0
What are you looking for?