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.

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 \[b=-6\]