Question

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

Answer 1

Answer:

Let p(x)=2x2−(3k+1)x−9

Let α,β be the roots of this polynomial.

We know that 

α+β=23k+1= Sum of the roots 

 

αβ=2−9= Products of the roots

Given that :  One zero is negative of the other 

⇒α=−β

⇒α+β=0

⇒α+β=23k+1=0

⇒3k+1=0

⇒k=−31

Answer 2

Note: the given question is incomplete as it should be as follows:

If one zero of the quadratic polynomial $\[39{y^2} - \left( {2k + 1} \right)y - 22\]$ is negative of the other, then find the value of k.

Given:

One zero of the given polynomial is negative of the other, i.e., $\[\alpha = - \beta \]$.

Polynomial: $\[39{y^2} - \left( {2k + 1} \right)y - 22\]$.

To find: the value of k.

Solution:

Note that from the question, $\[\alpha = - \beta \]$.

Compare the given quadratic polynomial with the standard form of the quadratic equation, $ax^{2}+bx+c$.

Therefore, $a=39, b=-(2k+1), c=-22$.

Know that,

Sum of the zeroes$\[ = - \frac{{\text{ coefficient of} x}}{{\text{coefficient\,of}\,{x^2}}}\]$

$\[ \Rightarrow \alpha + \beta = - \frac{b}{a}\]$

$\[ \Rightarrow \alpha + \beta = - \frac{{ - \left( {2k + 1} \right)}}{{39}}\]$

Substitute, $\[\alpha = - \beta \]$.

$\[ \Rightarrow - \beta + \beta = \frac{{\left( {2k + 1} \right)}}{{39}}\]$

$\[ \Rightarrow 0 = \frac{{\left( {2k + 1} \right)}}{{39}}\]$

$\[\begin{array}{l} \Rightarrow 2k + 1 = 39\\ \Rightarrow 2k = 39 - 1\\ \Rightarrow 2k = 38\end{array}\]$

$\[\begin{array}{l} \Rightarrow k = \frac{{38}}{2}\\ \Rightarrow k = 19\end{array}\]$

Therefore, the value of k is 19.

Hence, If one zero of the quadratic polynomial $\[39{y^2} - \left( {2k + 1} \right)y - 22\]$ is negative of the other, then the value of k is 19.