Q. y=-1/√5 ,y=2/√5 are zeroes of p(y) = (1-5y²)
Answers
quadratic polynomial is y2+235y−5.
The zeros of the polynomial are:
y2+235y−5=0
⟹ y2+25y−25y−5=0
⟹ y(y+25)−25(y+25)=0
⟹ (y+25)(y−25)=0
⟹ y−25 and y=25=0
The zeros are −25 and 2
Answer:
−2
5
,
2
5
The quadratic polynomial is y
2
+
2
3
5
y−5.
The zeros of the polynomial are:
y
2
+
2
3
5
y−5=0
⟹ y
2
+2
5
y−
2
5
y−5=0
⟹ y(y+2
5
)−
2
5
(y+2
5
)=0
⟹ (y+2
5
)(y−
2
5
)=0
⟹ y−2
5
and y=
2
5
=0
The zeros are −2
5
and
2
5
.
Now verifying the relation of zeros with the coefficients of the quadratic polynomial y
2
+
2
3
5
y−5:
Comparing it with the standard from of quadratic equation ax
2
+bx+c, we get, a=1, b=
2
3
5
and c=−5
Thus,
Sum of the zeros =−2
5
+
2
5
=
2
−4
5
+
5
=
2
−3
5
=−
a
b
.
Product of the zeros =−2
5
×
2
5
=−5
=
a
c
.
Hence, verified.
Step-by-step explanation:
use this method and find the answer