• question_answer If $x=\frac{2}{3}$ and $x=-3$ are roots of the quadratic equation $a{{x}^{2}}+7x+b=0$, find the values of a and b.

 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 $b=-6$