Question

If one zero of the quadratic polynomial p(x)=x^2-5x+k is -4, then the value of k is​

Answer 1

Answer:

Given that there is a quadratic polynomial,

p(x) = 5 {x}^{2} + 13x - kp(x)=5x

2

+13x−k

It's also given that the zeroes are reciprocal of each other.

Let, one of the zeroes is \bold{\alpha}α

Therefore, other zero will be \bold{\dfrac{1}{\alpha}}

α

1

Now, we know that,

Sum of zeroes =- \dfrac{coeff.\; of\; x}{coeff.\;of\;{x}^{2}}−

coeff.ofx

2

coeff.ofx

Product of zeroes = \dfrac{constant\;term}{coeff.\;of\;{x}^{2}}

coeff.ofx

2

constantterm

Therefore, we will get,

\begin{gathered} = > \alpha \times \dfrac{1}{ \alpha } = \frac{ - k}{5} \\ \\ = > \dfrac{ - k}{5} = 1 \\ \\ = > - k = 5 \times 1 \\ \\ = > - k = 5 \\ \\ = > k = - 5\end{gathered}

=>α×

α

1

=

5

−k

=>

5

−k

=1

=>−k=5×1

=>−k=5

=>k=−5

Hence, the required value of k = -5