Question

frnds pls answer this question

maths class 10 polynomials

no spamming pls

answer asap​

Answer 1

Solution:

Here I am using m,n instead of

Alfa and Beta.

Given m,n are zeroes of f(x)=x²-3x-2

Compare f(x) with ax²+bx+c,

we get

a = 1 , b = -3 , c=-2

i) m+n = -b/a = -(-3)/1=-3

ii) mn = c/a = (-2)/1 = -2

iii) m²+n² =(m+n)²-2mn

= (-3)²-2(-2)

= 9+4

= 13

Now,

If 1/(2m+n ) , 1/(2n+m)

are two zeroes ,

iv ) 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)]/[5mn+2(m²+n²)]

= [3×(-3)]/[5(-2)+2×13]

= (-9)/[-10+26]

= (-9)/16

v) Product of the zeroes

= [1/(2m+n)][1/(2n+m)]

= 1/[(2m+n)(2n+m)]

= 1/[4mn+2m²+2n²+mn]

= 1/[5mn+2(m²+n²)]

= 1/[5(-2)+2×13]

= 1/(-10+26)

= 1/16

______________________

Form of quadratic polynomial

whose zeroes are p,q is

k[x²-(p+q)x+pq]

_____________________

Here,

Required polynomial is

k[ x² -(-9/16)x + (1/16)]

For real values of k , it is true.

Let k = 16, then

the polynomial is

16[ x²-(-9/16)x+1/16]

= 16x² +9x+1

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