find zeroes of the following
quadratic
polynomial and
and verify
relation
i) 2x^2+3x+1
Answers
Answer:
Zeroes of the polynomial are 1 and
2
1
\sf\orange{Given:}Given:
\sf{The \ given \ quadratic \ polynomial \ is}The given quadratic polynomial is
\sf{\implies{2x^{2}-3x+1}}⟹2x
2
−3x+1
\sf\pink{To \ find:}To find:
\sf{Zeroes \ of \ the \ polynomial. }Zeroes of the polynomial.
\sf\green{\underline{\underline{Solution:}}}
Solution:
\sf{The \ given \ quadratic \ polynomial \ is}The given quadratic polynomial is
\sf{\implies{2x^{2}-3x+1}}⟹2x
2
−3x+1
\sf{\implies{2x^{2}-2x-x+1}}⟹2x
2
−2x−x+1
\sf{\implies{2x(x-1)-1(x-1)}}⟹2x(x−1)−1(x−1)
\sf{\implies{(x-1)(2x-1)}}⟹(x−1)(2x−1)
\sf{\implies{\therefore{x=1 \ or \ \frac{1}{2}}}}⟹∴x=1 or
2
1
\sf{Zeroes \ of \ the \ polynomial \ are \ 1 \ and \ \frac{1}{2}}Zeroes of the polynomial are 1 and
2
1
_______________________________________
\blue{\underline{\underline{Verification:}}}
Verification:
\sf{The \ given \ quadratic \ polynomial \ is}The given quadratic polynomial is
\sf{\implies{2x^{2}-3x+1}}⟹2x
2
−3x+1
\sf{Here, \ a=2, b=-3 \ and \ c=1}Here, a=2,b=−3 and c=1
\sf{Let \ \alpha \ be \ 1 \ and \ \beta \ be \ \frac{1}{2}}Let α be 1 and β be
2
1
_________________________
\sf{\alpha+\beta=1+\frac{1}{2}}α+β=1+
2
1
\sf{\therefore{\alpha+\beta=\frac{3}{2}...(1)}}∴α+β=
2
3
...(1)
\sf{\frac{-b}{a}=\frac{-(-3)}{2}}
a
−b
=
2
−(−3)
\sf{\therefore{\frac{-b}{a}=\frac{3}{2}...(2)}}∴
a
−b
=
2
3
...(2)
\sf{...from \ (1) \ and \ (2)}...from (1) and (2)
\sf{Sum \ of \ zeroes=\frac{-b}{a}}Sum of zeroes=
a
−b
_________________________
\sf{\alpha\beta=1\times\frac{1}{2}}αβ=1×
2
1
\sf{\therefore{\alpha\beta=\frac{1}{2}...(3)}}∴αβ=
2
1
...(3)
\sf{\frac{c}{a}=\frac{1}{2}...(4)}
a
c
=
2
1
...(4)
\sf{...from \ (3) \ and \ (4)}...from (3) and (4)
\sf{Product \ of \ zeroes=\frac{c}{a}}Product of zeroes=
a
c
\sf{Hence, \ verified. }Hence, verified.
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Answer:
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