10th Class Mathematics Solved Paper - Mathematics-2016 Delhi Term-II Set-I

  • 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 \[b=-6\]


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