If x=2/3 and x=-3are roots of the quadratic equation ax+7x+b=0 find the values of a and b

If x=2/3 and x=-3are roots of the quadratic equation ax+7x+b=0 find th

1.	If x=2/3 and x=-3are roots of the quadratic equation ax+7x+b=0 find the values of a and b
Answer» {tex}\\because{/tex}\xa0x =\xa0{tex}\\frac{2}{3}{/tex}\xa0is a root of {tex}ax^2 + 7x + b = 0{/tex}{tex}\\therefore{/tex}\xa0a({tex}\\frac{2}{3}{/tex})2 + 7{tex}\\times{/tex}{tex}\\frac{2}{3}{/tex}\xa0+ b = 0{tex}\\Rightarrow{/tex}{tex}\\frac{4a + 42 + 9b}{9}{/tex}\xa0= 0\xa0{tex}\\Rightarrow{/tex}\xa0{tex}4a + 9b + 42 = 0{/tex} ...(i)Also x = 3 is a root{tex}\\therefore{/tex}\xa0{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}\xa077a = -147\xa0{tex}\\Rightarrow{/tex}\xa0a =\xa0{tex}\\frac{-147}{77}{/tex}\xa0=\xa0{tex}\\frac{-21}{11}{/tex}When a =\xa0{tex}\\frac{-21}{11}{/tex}, eq.(i) becomes-4{tex}\\times{/tex}{tex}\\frac{21}{11}{/tex}\xa0+ 9b + 42 = 0{tex}\\Rightarrow{/tex}{tex}\\frac { - 84 + 99 b + 462 } { 11 }{/tex}\xa0= 0{tex}\\Rightarrow{/tex}99b + 378 = 0 b =\xa0{tex}\\frac{-378}{99}{/tex}\xa0=\xa0{tex}\\frac{-42}{11}{/tex}{tex}\\therefore{/tex}\xa0a =\xa0{tex}\\frac{-21}{11}{/tex}, b =\xa0{tex}\\frac{-42}{11}{/tex}.