Question

solve proper.......​

Answer 1

ANSWER:-

Given:

p(x)= 2x³ + 5x² -9x -18

&

The product of two zeroes is 3.

To find:

Find the other zeroes.

Solution:

$</u><u>A</u><u>ssum e</u><u>,</u><u> \: \alpha \beta = - 3...........(1) \\ \\ = > \alpha \beta \gamma = - \frac{d}{a} = \frac{ - ( - 18)}{2} \\ \\ = > \alpha \beta \gamma = \frac{18}{2} = 9 \\ so \\ = > ( - 3)y = 9 \\ = > - 3y = 9 \\ = > y = - \frac{9}{3} \\ = > y = - 3 \\ </u><u>T</u><u>herefore </u><u>,</u><u>\\ = > \alpha + \beta + \gamma = \frac{ - 5}{2} \\ \\ = > \alpha + \beta - 3 = \frac{ - 5}{2} \\ \\ = > \alpha + \beta = 3 - \frac{5}{2} \\ \\ = > \alpha + \beta = \frac{6 - 5}{2} \\ \\ = > \alpha + \beta = \frac{1}{2} \\ = > \alpha = (\frac{1}{2} - \beta ) \\ \\ = > ( \frac{1}{2} - \beta ) \beta = - 3 \: \: \: \: \: \: \: \: </u><u>[</u><u>P</u><u>utting</u><u> \: in \: eq.1</u><u>]</u><u> \\ \\ = > \frac{ \beta }{2} - { \beta }^{2} = - 3 \\ \\ = > \beta - 2 { \beta }^{2} = - 6 \\ \\ = > \beta - 2 { \beta }^{2} + 6 = 0 \\ \\ = > 2 { \beta }^{2} - \beta - 6 = 0 \\ \\ = > 2 { \beta }^{2} - 4 \beta + 3 \beta - 6 = 0 \\ \\ = > 2 \beta ( \beta - 2) + 3( \beta - 2) = 0 \\ \\ = > ( \beta - 2)(2 \beta + 3) = 0 \\ \\ = > \beta - 2 = 0 \: \: \: or \: \: \: 2 \beta + 3 = 0 \\ \\ = > \beta = 2 \: \: \: or \: \: \beta = - \frac{3}{2}$

So,

⚫One zero, alpha= 2

⚫Other zero, Beta= -3/2

Option (1)✓

Hope it helps ☺️