So, the value of x2
+ 7x + 10 is zero when x + 2 = 0 or x + 5 = 0,
i.e., when x = –2 or x = –5.
Therefore, the zeroes of x2
+ 7x + 10 are –2 and –5.
Now, sum of the zeroes= –2 + (–5) = – (7) =
-(7)/
1
=
-(coefficient of )/
coefficient of
x2
Product of the zeroes = –2 × (–5) = 10 =
10/
1
=
constant term/
coefficient of x2
Example-4. Find the zeroes of the polynomial x
2
– 3 and verify the relationship between the
zeroes and the coefficients.
Solution : Recall the identity a2
– b2
= (a – b) (a + b).
Using it, we can write:
x
2– 3 = (x – √3 ) (x + √3 )
So, the value of x
2 – 3 is zero when x = √3 or x = – √3 .
Therefore, the zeroes of x
2 – 3 are √3 and – √3 .
Sum of the zeroes = (√3)+ √(– 3 ) = 0 = 2
(coefficient of )/
coefficient of
x2
Product of zeroes = ( √3 ) × (– √3 ) = – 3 =
-3/
1
=
constant term/
coefficient of x2
Example-5. Find the quadratic polynomial, whose sum and product of the zeroes are – 3 and
2, respectively.
Solution : Let the quadratic polynomial be ax2
+ bx + c, and its zeroes be a and b. We have
α+β = – 3 =
-b/
a
,
and αβ = 2 =
c/
a
If we take a = 1, then b = 3 and c = 2
So, one quadratic polynomial which fits the given conditions is x
2
+ 3x + 2.
page no:67
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