Question

22. If one zero of the polynomial (k + 1)x2 – 5x + 5 is multiplicative inverse of the
other ,find the zeroes of kx2 – 3kx + 9, where k is constant.

Answer 1

ANSWER:-

Given:

If one zero of the polynomial (k+1)x²-5x+5 is multiplicative inverse of the other.

To find:

Find the zeroes of kx² -3kx +9, where k is constant.

Solution:

We have,

(k+1)x² -5x +5.

Zeroes of the polynomial

$= > \alpha , \frac{1}{ \alpha } \: \: \: \: \: \: [multiplicative \: inverse]$

Product of zeroes

$= > \frac{c}{a} = \frac{5}{k + 1} \\ so \\ = > \alpha \times \frac{1}{ \alpha } = \frac{5}{k + 1} \\ \\ = > k + 1 = 5 \\ \\ = > k = 5 - 1 \\ \\ = > k = 4$

Therefore,

Putting the value of k in given polynomial;

kx² -3kx +9

=) 4x² -3(4)x +9

=) 4x² -12x +9=0

=) 4x² -6x -6x +9=0

=) 2x(2x -3) -3(2x -3)=0

=) (2x -3) (2x-3)=0

=) 2x -3=0. OR 2x-3 =0

=) 2x = 3 OR 2x= 3

=) x= 3/2 OR x= 3/2

So,

$\alpha = \frac{3}{2} \: \: and \: \: \beta = \frac{3}{2}$

Hope it helps ☺️