find the zero by relationship
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Step-by-step explanation:
.3x
2
+4x−4
⇒3x
2
+6x−2x−4
3x(x+2)−2(x+2)
(3x−2)(x+2)
⇒ Zeroes : x=
3
2
,−2
Sum of zeroes =
3
2
+(−2)=
3
−4
=
a
−b
Product of zeroes =(
3
2
)(−2)=
3
−4
=
a
c
The sum of zeroes and product of zeroes are equal to
a
−b
&
a
c
respectively.
ii).21t
2
−t−2
⇒21t
2
−7t+6t−2
7t(3t−1)+2(3t−1)
(7t+2)(3t−1)
⇒ Zeroes :t=
7
−2
&
3
1
Sum of zeroes =
7
−2
+
3
1
=
21
1
=
a
−b
Product of zeroes =(
7
−2
)(
7
−2
)=(
3
1
)=
a
c
verified.
iii).2u
2
−7
⇒u=
2
7
,−
2
7
⇒ Sum of zeroes =0=
a
−b
Product of zeroes =
2
−7
=
a
c
verified.
Hence, solved.
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