If x=2/3 and x=-3 are roots …

{tex}\because{/tex} x = {tex}\frac{2}{3}{/tex} is a root of {tex}ax^2 + 7x + b = 0{/tex}
{tex}\therefore{/tex} a({tex}\frac{2}{3}{/tex})² + 7{tex}\times{/tex}{tex}\frac{2}{3}{/tex} + b = 0
{tex}\Rightarrow{/tex}{tex}\frac{4a + 42 + 9b}{9}{/tex} = 0 {tex}\Rightarrow{/tex} {tex}4a + 9b + 42 = 0{/tex} ...(i)
Also x = 3 is a root
{tex}\therefore{/tex} {tex}a(3)^2 + 7\times3 + b = 0{/tex}
{tex}\Rightarrow{/tex}{tex}9a + b + 21 = 0{/tex}
{tex}\Rightarrow{/tex}{tex}9(9a + b + 21) = 9\times0{/tex}
{tex}\Rightarrow{/tex}{tex}81a + 9b + 189 = 0 {/tex}...(ii)
(ii) and (i), we get

{tex}\Rightarrow{/tex} 77a = -147 {tex}\Rightarrow{/tex} a = {tex}\frac{-147}{77}{/tex} = {tex}\frac{-21}{11}{/tex}
When a = {tex}\frac{-21}{11}{/tex}, eq.(i) becomes
-4{tex}\times{/tex}{tex}\frac{21}{11}{/tex} + 9b + 42 = 0
{tex}\Rightarrow{/tex}{tex}\frac { - 84 + 99 b + 462 } { 11 }{/tex} = 0
{tex}\Rightarrow{/tex}99b + 378 = 0 b = {tex}\frac{-378}{99}{/tex} = {tex}\frac{-42}{11}{/tex}
{tex}\therefore{/tex} a = {tex}\frac{-21}{11}{/tex}, b = {tex}\frac{-42}{11}{/tex}.