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If alfa and bita are the zeros of polynomial 2x^2-5x+7,find a polynomial whose zeros are
2alfa+3 and 2bita+3.
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Answers
Answered by
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p(x) = 2x²-5x+7
Here , @+ß=-(-5)/2 =5/2
@ß= 7/2
Sum of zeroes of f(x) [S] = 2@+3+2ß+3=2(@+ß)+6 = 2(5/2)+6 = 11
Product of zeroes of f(x) [P] = (2@+3)(2ß+3) = 4@ß +6(@+ß)+9= 4(7/2) + 6(5/2) + 9 = 14+15+9=38
f(x) = x²-Sx+P
f(x) = x²-11x+38
Here , @+ß=-(-5)/2 =5/2
@ß= 7/2
Sum of zeroes of f(x) [S] = 2@+3+2ß+3=2(@+ß)+6 = 2(5/2)+6 = 11
Product of zeroes of f(x) [P] = (2@+3)(2ß+3) = 4@ß +6(@+ß)+9= 4(7/2) + 6(5/2) + 9 = 14+15+9=38
f(x) = x²-Sx+P
f(x) = x²-11x+38
Answered by
1
2x^2-5x+7
a=2, b=-5 c= 7
α+β=-b/a= -(-5/2)=5/2
α β= c/a= 7/2
sum of zeroes=
2α+3 + 2β +3 =( 2α+ 2β )+3+3
2(α +β)+(3+3)
2(α +β)+6)
2(5/2)+6)
2(5/2)+6
10/2+6
5+6
11
product of zeroes
(2α+3 )( 2β +3)
4αβ+6α+6β+9
4αβ+6(α+β)+9
4×7/2+6(5/2)+9
14+15+9
38
polynomial= x^2-(α+β)x+α β
=x^2-11x+38
a=2, b=-5 c= 7
α+β=-b/a= -(-5/2)=5/2
α β= c/a= 7/2
sum of zeroes=
2α+3 + 2β +3 =( 2α+ 2β )+3+3
2(α +β)+(3+3)
2(α +β)+6)
2(5/2)+6)
2(5/2)+6
10/2+6
5+6
11
product of zeroes
(2α+3 )( 2β +3)
4αβ+6α+6β+9
4αβ+6(α+β)+9
4×7/2+6(5/2)+9
14+15+9
38
polynomial= x^2-(α+β)x+α β
=x^2-11x+38
Thank You
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