find the zeros of the given polynomial and verify the relationship between zeroes and co-fficent x2-20x+91
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Answered by
2
x²-13x-7x +91
x(x-13)-7(x-13)
(x-13)(x-7)
x=13, x=7
α+β = 13+7 = 20
α+β= -b/a
=-(-20)/1
=20
αβ = -13*-7
=91
αβ = c/a
=91
x(x-13)-7(x-13)
(x-13)(x-7)
x=13, x=7
α+β = 13+7 = 20
α+β= -b/a
=-(-20)/1
=20
αβ = -13*-7
=91
αβ = c/a
=91
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Answered by
0
Lets solve it by doing SPLITTING OF MIDDLE TERM:
x²-20+91
x²-7x-13x+91
x(x-7)-13(x-7)
(x-13)(x-7)=0
x-13=0 x-7=0
x=13 x=7
SUM OF ZEROES:
13+7=20
BY FORMULA:
α+β=-b÷a
= -(-20)÷1
=20
PRODUCT OF ZEROES:
13×7=91
BY FORMULA:
αβ=c÷a
=91÷1
=91
HENCE SOLVED
x²-20+91
x²-7x-13x+91
x(x-7)-13(x-7)
(x-13)(x-7)=0
x-13=0 x-7=0
x=13 x=7
SUM OF ZEROES:
13+7=20
BY FORMULA:
α+β=-b÷a
= -(-20)÷1
=20
PRODUCT OF ZEROES:
13×7=91
BY FORMULA:
αβ=c÷a
=91÷1
=91
HENCE SOLVED
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