Question

find the sum of zeros of p(x) = x^2 - 2020x + 2021​

Answer 1

Given the polynomial x

2

−3x−5

For solving we have to make it a zero,

∴x

2

−3x−5=0

Now, by formula,

x=

2⋅1

−(−3)±

(−3)

2

−4⋅1⋅(−5)

=

2

3±

9+20

=

2

3±

29

∴x=

2

3+

29

,

2

3−

29

∴ Sum of the roots, (S)=

2

3+

29

+

2

3−

29

=

2

3+

29

+3−

29

=

2

6

=3

and product of the roots, (P)=(

2

3+

29

)×(

2

3−

29

)

=

4

9−29

=

4

−20

=−5

Now, using S and P as zeros we have,

x=3

⇒x−3=0

and x=−5

⇒x+5=0

∴ the required polynomial is,

(x−3)(x+5)

=x

2

+2x−15.

Answer 2

Answer:

0.995

Step-by-step explanation:

$P(x)=x^2-2020x+2021$

We know that,

sum of zeroes = $-\dfrac{(\text{coefficient of }x)}{(\text{coefficient of }x^2)}$

$=-\dfrac{-2020}{2021}$

$=\dfrac{2020}{2021}$

$=0.995 \text{ (approx.)}$