Question

find the value of k so that the sum of the zeros of the polynomial 3x^2+(2k+1)x-k+5​

Answer 1

Answer:

Let P(x)=3x

2

+2kx+x−k−5

=3x

2

+x(2k+1)−(k+5)

i.e; ax

2

+bx+c

a=3;b=2k+1;c=−(k+5)

1) Sum of zeroes =

a

−b

=

3

−(2k+1)

2) Product of zeroes =

a

c

=

3

−(k+5)

According to given

−(2k+1)/3=

2

1

[−(k+5)/3]

2(2k+1)=k+5

4k+2=k+5

3k=3⇒k=1