Question

If one zero of the quadratic polynomial 39y² - (2k + 1)y - 22 is negative of the other, then value of k is​

Answer 1

Answer:

Here is your answer..

HOPE IT HELPS YOU..

Answer 2

Answer:

The value of k is $k=-\frac{1}{2}$

Step-by-step explanation:

The given quadratic polynomial is $39y^2-(2k+1)y-22$ and it is also given that on zero of the quadratic polynomial is negative of the other.

Let $y_{1}$ and $y_{2}$ be the zeroes of the given polynomial.

$= > y_{1}=-y_{2}$

We know that for a quadratic polynomial $ax^2+bx+c$, the sum of its zeroes is $-\frac{b}{a}$

Therefore,

For our polynomial we get

$y_{1}+y_{2}=\frac{2k+1}{39} \\= > -y_{2}+y_{2}=\frac{2k+1}{39}\\= > 2k+1=0= > k=-\frac{1}{2}$

Therefore,

The value of k is $k=-\frac{1}{2}$

#SPJ3