Question

if one zero of the quadratic polynomial (k2+9)X2 +13x+6k) is reciprocal of the other find k

Answer 1

In this question I have used alpha and reciprocal of alpha.

Answer 2

Answer:

The value of k is 3.

Step-by-step explanation:

The given quadratic polynomial is

$(k^2+9)x^2+13x+6k$

If α and β are the roots of quadratic equation

$ax^2+bx+c$

then

$\alpha \beta =\frac{c}{a}$

It is given that one zero of the quadratic polynomial is reciprocal of the other.

Let the roots be p and 1/p.

$p\times \frac{1}{p}=\frac{6k}{k^2+9}$

$1=\frac{6k}{k^2+9}$

$k^2+9=6x$

$k^2-6x+9=0$

$(k-3)^2=0$

$k-3=0$

$k=3$

Therefore the value of k is 3.