if X and B are the zeros of the qaudratic polynomial f(x)=x2-3x+2,then find x2+b2
Answers
Answer:Given:
If m and n are the zeroes of
quadratic polynomial f(x)=x²-3x-2
To find :
Quadratic polynomial whose
zeroes are 1/(2m+n) and 1/(2n+m) .
Explanation:
Compare f(x)= x²-3x-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) m²+n² = (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(m²+n²)+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(m²+n²) + 5mn ]
= 1/[ 2×13 + 5(-2)]
= 1/(26-10)
= 1/16 ---(5)
______________________
Form of a quadratic polynomial
is
k[x²-(sum of the zeroes)x+product of the zeroes]
______________________
Here ,
Required polynomial is
k[ x²-(9/16)x+1/16]
For all real values of k it is true.
let k = 16,
16[x²-(9/16)x+1/16]
= 16x²-9x+1
Therefore,
Polynomial whose zeroes are
1/(2m+n) and 1/(2n+m) is
16x²-9x+1
hope it helps u
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Step-by-step explanation: