find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficients 3.3
Answers
Answer:
bnnkk
Step-by-step explanation:
jzhzhsjzjjsiskdndjdmhxkdbd
Answer:
ok
Step-by-step explanation:
x
2
−2x−8
Factorize the equation, we get (x+2)(x−4)
So, the value of x
2
−2x−8 is zero when x+2=0,x−4=0, i.e., when x=−2 or x=4.
Therefore, the zeros of x
2
−2x−8 are -2 and 4.
Now,
⇒Sum of zeroes =−2+4=2=−
1
2
=−
Coefficient of x
2
Coefficient of x
⇒Product of zeros =(−2)×(4)=−8 =
1
−8
=
Coefficient of x
2
Constant term
(ii) 4s
2
−4s+1
Factorize the equation, we get(2s−1)(2s−1)
So, the value of 4s
2
−4s+1 is zero when 2s−1=0,2s−1=0, i.e., when s=
2
1
or s=
2
1
.
Therefore, the zeros of 4s
2
−4s+1 are
2
1
and
2
1
.
Now,
⇒Sum of zeroes =
2
1
+
2
1
=1=−
4
−4
=−
Coefficient of s
2
Coefficient of s
⇒Product of zeros =
2
1
×
2
1
=
4
1
=
4
1
=
Coefficient of s
2
Constant term
(iii) 6x
2
−3−7x
Factorize the equation, we get (3x+1)(2x−3)
So, the value of 6x
2
−3−7x is zero when 3x+1=0,2x−3=0, i.e., when x=−
3
1
or x=
2
3
.
Therefore, the zeros of 6x
2
−3−7x are −
3
1
and
2
3
.
Now,
⇒Sum of zeroes = −
3
1
+
2
3
=
6
7
=−
6
−7
=−
Coefficient of x
2
Coefficient of x
⇒Product of zeros = −
3
1
×
2
3
=−
2
1
=
6
−3
=
2
−1
=
Coefficient of x
2
Constant term
(iv) 4u
2
+8u
Factorize the equation, we get 4u(u+2)
So, the value of 4u
2
+8u is zero when 4u=0,u+2=0, i.e., when u=0 or u=−2.
Therefore, the zeros of 4u
2
+8u are 0 and −2.
Now,
⇒Sum of zeroes = 0−2=−2=−
4
8
=−2=−
Coefficient of u
2
Coefficient of u
⇒Product of zeros = −0x−2=0=
4
0
=0=
Coefficient of u
2
Constant term