Answer:
The given polynomial is, \[p(x)=a{{x}^{2}}+7x+b\] \[\therefore p\left( \frac{2}{3} \right)=a{{\left( \frac{2}{3} \right)}^{2}}+7\left( \frac{2}{3} \right)+b=0\] \[=\frac{4a}{9}+\frac{14}{3}+b=0\] ?(i) and, \[p(-3)=a{{(-3)}^{2}}+7(-3)+b=0\] \[\Rightarrow 9a-21+b=0\] ?(ii) Solving equation (i) and (ii), we get \[_{\begin{smallmatrix} 81a\,-\,189\,+\,9b\,=\,0 \\ -\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \overline{\,\,\,\,\,\,\,-\,\,77a\,+\,231\,=\,0} \end{smallmatrix}}^{4a\,\,\,+\,\,\,42\,\,+\,\,9b\,=\,0}\] \[a=\frac{231}{77}=3\] Putting \[a=3\text{ }m\]eq. (ii) we get, \[9(3)-21+b=0\] \[\Rightarrow b=-6\] \[\therefore a=3\] and \[b=-6\]
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