Question

One zero of the polynomial 6x 2 -px -3 is 3/2. The zeroes of the polynomial px2+x+k are equal . The value of k is

a. -1/14
b. 1/28
c. 1/7
d. 7​

Answer 1

Given: One of the zroes of the polynomial $\[6{x^2} - px - 3\]$ is $\[\frac{3}{2}\]$.

The zeroes of the polynomial $\[p{x^2} + x + k\]$are equal.

To find: the value of k.

Solution:

Let the polynomial be $\[q\left( x \right)\]$, then

$\[\begin{array}{l}q\left( x \right) = 6{x^2} - px - 3\\q\left( {\frac{3}{2}} \right) = 0\end{array}\]$

Find the value of p.

$\[ \Rightarrow 6 \times \left( {\frac{9}{4}} \right) - \left( {\frac{3}{2}} \right)p - 3 = 0\]$

$\[\begin{array}{l} \Rightarrow \frac{{27}}{2} - \frac{3}{2}p = 3\\ \Rightarrow \frac{3}{2}p = \frac{{27}}{2} - 3\end{array}\]$

$\[ \Rightarrow \frac{3}{2}p = \frac{{27 - 6}}{2}\]$

$\[\begin{array}{l} \Rightarrow \frac{3}{2}p = \frac{{21}}{2}\\ \Rightarrow p = 7\end{array}\]$

Given that, The zeroes of the second polynomial are equal.

$\[\begin{array}{l}p{x^2} + x + k = 0\\ \Rightarrow 7{x^2} + x + k = 0\end{array}\]$

Assume that both the zeroes is $l$.

$\[{2l = \left( { - \frac{b}{a}} \right) = - \frac{1}{7} - - - - - - \left( 1 \right)}\]$

$\[{{l^2} = \left( {\frac{c}{a}} \right) = \left( {\frac{k}{7}} \right){\rm{ - - - - - - }}\left( 2 \right)}\]$

Divide equation (2) by (1).

$\[{ \Rightarrow \frac{l}{2} = \left( {\frac{k}{2}} \right) \times \left( { - 7} \right)}\]$

$\[{ \Rightarrow \frac{l}{2} = - k}\]$

$\[{ \Rightarrow l = - 2k}\]$

Find the value of k.

$\[{m\left( l \right) = 0}\]$

$\[{ \Rightarrow m\left( { - 2k} \right){\rm{ }} = {\rm{ }}0}\]$

$\[{ \Rightarrow 7{{\left( { - 2k} \right)}^2} + \left( { - 2k} \right) + k{\rm{ }} = 0}\]$

$\[{ \Rightarrow 28{k^2} - k = 0}\]$

$\[{ \Rightarrow k\left( {28k - 1} \right) = 0}\]$

$\[\begin{array}{l} \Rightarrow k = 0\,or\,28k - 1 = 0\\ \Rightarrow k = 0\,or\,k = \frac{1}{{28}}\end{array}\]$

Therefore, the value of k is $\[\frac{1}{{28}}\]$.

Hence, the correct answer is option (b). i.e., $\[\frac{1}{{28}}\]$.

Answer 2

Answer:

b. 1/28

Step-by-step explanation: