If the zeros of the polynomial …

Let f(x) = 2x²- 5x - 3
Let the zeroes of polynomial be {tex}\alpha{/tex} and {tex}\beta{/tex}, then
Sum of zeroes = {tex}\alpha + \beta = \frac { 5 } { 2 }{/tex}
Product of zeroes = {tex}{/tex}{tex}\alpha \beta = - \frac { 3 } { 2 }{/tex}
According to the question, zeroes of x²+ px + q are {tex}2 \alpha \text { and } 2 \beta{/tex}.
{tex}\therefore\ {/tex} Sum of zeroes = {tex}{/tex}{tex}- \frac { \text { Coeff. of } x } { \text { Coeff of } x ^ { 2 } } = \frac { - p } { 1 }{/tex}
{tex}\Rightarrow{/tex} -p = {tex}2 \alpha + 2 \beta = 2 ( \alpha + \beta ){/tex}
{tex}{/tex}{tex}\Rightarrow\ {/tex} -p= {tex}2 \times \frac { 5 } { 2 } = 5{/tex} or p = -5
Now ,Product of zeroes = {tex}\frac { \text { Constant term } } { \text { Coeff of } x ^ { 2 } } = \frac { q } { 1 }{/tex}
{tex}\Rightarrow{/tex} q = {tex}2 \alpha \times 2 \beta = 4 \alpha \beta{/tex}
{tex}\Rightarrow{/tex} q = {tex}4 \left( - \frac { 3 } { 2 } \right){/tex} = -6
{tex}\therefore{/tex} p = -5 and q = -6