find the zeros of the following quadratic polynomials and verify the relation between the zeros and the coefficient. 2a^2-2√2a+1 About the author Harper
Step-by-step explanation: f(x)=x 2 −2x−8 =(x−4)(x+2) Zeroes: −2,4 Sum of zeroes: −2+4=2 Product of zeroes:−2(4)=−8 ii) g(s)=4s 2 −4s+1 =(2s−1) 2 Zeroes: 2 1 , 2 1 Sum of zeroes : 2 1 + 2 1 = a −b =− 4 (−4) =1 Product of zeroes: 2 1 ⋅ 2 1 = a c = 4 1 ii)h(t)=t 2 −15=(t− 1 5)(t+ 1 5) Sum of zeroes: 1 5+(− 1 5)=0 Product of zeroes: ( 1 5)(− 1 5)=−15= a c iv) 6x 2 −3−7x Zeroes : 2 3 , 3 −1 Sum of zeroes: 2 3 − 3 1 = 6 7 = a −b Product of zeroes: 2 3 3 −1 = 6 −3 = a c v) p(x)=x 2 +2 2 −6 =(x− 6−2 2 )(x+ 6−2 2 ) Zeroes: − 6−2 2 , 6−2 2 Sum of zeroes: 6−2 2 − 6−2 2 =0= a −b Product of zeroes: ( 6−2 2 )(− 6−2 2 )=−6+2 2 = a c vi) q(x)= 3 x 2 +10x+7 3 Zeroes:− 3 , 3 −7 Sum of zeroes: −( 3 + 3 7 )= 3 −10 = a −b Product of zeroes: (− 3 )(− 3 7 )=7= 3 7 3 = a c vii) f(x)=x 2 −( 3 +1)x+ 3 Zeroes: 1, 3 Sum of zeroes: 1+ 3 = a c Product of zeroes: 1× 3 = 3 = a c viii)g(x)=a(x 2 +1)−x(a 2 +1)=ax 2 −(a 2 +1)x+a Zeroes: a, a 1 Sum of zeroes: a+ a 1 = a (a 2 +1) = c ′ b ′ Product of zeroes: a× a 1 =1= a a = a ′ c ′ Reply
Step-by-step explanation:
f(x)=x
2
−2x−8
=(x−4)(x+2)
Zeroes: −2,4
Sum of zeroes: −2+4=2
Product of zeroes:−2(4)=−8
ii) g(s)=4s
2
−4s+1
=(2s−1)
2
Zeroes:
2
1
,
2
1
Sum of zeroes :
2
1
+
2
1
=
a
−b
=−
4
(−4)
=1
Product of zeroes:
2
1
⋅
2
1
=
a
c
=
4
1
ii)h(t)=t
2
−15=(t−
1
5)(t+
1
5)
Sum of zeroes:
1
5+(−
1
5)=0
Product of zeroes: (
1
5)(−
1
5)=−15=
a
c
iv) 6x
2
−3−7x
Zeroes :
2
3
,
3
−1
Sum of zeroes:
2
3
−
3
1
=
6
7
=
a
−b
Product of zeroes:
2
3
3
−1
=
6
−3
=
a
c
v) p(x)=x
2
+2
2
−6
=(x−
6−2
2
)(x+
6−2
2
)
Zeroes: −
6−2
2
,
6−2
2
Sum of zeroes:
6−2
2
−
6−2
2
=0=
a
−b
Product of zeroes: (
6−2
2
)(−
6−2
2
)=−6+2
2
=
a
c
vi) q(x)=
3
x
2
+10x+7
3
Zeroes:−
3
,
3
−7
Sum of zeroes: −(
3
+
3
7
)=
3
−10
=
a
−b
Product of zeroes: (−
3
)(−
3
7
)=7=
3
7
3
=
a
c
vii) f(x)=x
2
−(
3
+1)x+
3
Zeroes: 1,
3
Sum of zeroes: 1+
3
=
a
c
Product of zeroes: 1×
3
=
3
=
a
c
viii)g(x)=a(x
2
+1)−x(a
2
+1)=ax
2
−(a
2
+1)x+a
Zeroes: a,
a
1
Sum of zeroes: a+
a
1
=
a
(a
2
+1)
=
c
′
b
′
Product of zeroes: a×
a
1
=1=
a
a
=
a
′
c
′