Question

If one zero of the quadratic polynomial p (x)=4x2+8kx+8x-9 is negative of the other then find zeroes of kx2+3kx+2

Answer 1

Let the zeros be a
& -a

4x²+8kx+8x-9

4x²+(8k+8)x-9

sum of zeros=
$= \frac{ - coefficent \: of \: x}{coefficent \: of \: {x}^{2} }$


a+(-a)=-(8k+8)/4

0=-8k-8

8=-8k

-1=k

So,we get the value of k =-1
putting value of k in second equation

kx²+3kx+2

=> -x²+3×(-1)x+2

=> -x²-3x+2

Now,

-x²-3x+2

D=b²-4ac

=(3)²-4×(-1)×2

=9+8

=17

x=(-b±√D)/2a

=(3±√17)/-2

zeros of second equation

x=(3+√17)/-2

x=(3-√17)/-2


@Altaf

Answer 2

Step-by-step explanation:

Answer :-

→ k = 0 .

Step-by-step explanation :-

It is given that,

→ One zeros of the given polynomial is negative of the other .

Let one zero of the given polynomial be x .

Then, the other zero is -x .

•°• Sum of zeros = x + ( - x ) = 0 .

But, Sum of zeros = -( coefficient of x )/( coefficient of x² ) = - ( -8k )/4 .

==> 2k = 0 .

==> k = 0/2 .

•°• k = 0 .

Hence, it is solved.