10th Class Mathematics Solved Paper - Mathematics-2018

  • question_answer Find all zeroes of the polynomial \[(2{{x}^{4}}-9{{x}^{3}}+5{{x}^{2}}+3x-1)\]  if two of its zeroes are \[\left( 2+\sqrt{3} \right)\] and \[\left( 2-\sqrt{3} \right)\]

    Answer:

    Here, \[p\left( x \right)=2{{x}^{4}}-9{{x}^{3}}+5{{x}^{2}}+3x-1\]
    And two of its zeroes are \[\left( 2+\sqrt{3} \right)\] and \[\left( 2-\sqrt{3} \right)\]
    Quadratic polynomial with zeroes is given by,
                \[\left\{ x-\left( 2+\sqrt{3} \right) \right\}.\left\{ x-\left( 2-\sqrt{3} \right) \right\}\]
    \[\Rightarrow \]   \[\left( x-2-\sqrt{3} \right)\left( x-2+\sqrt{3} \right)\]
    \[\Rightarrow \]   \[{{(x-2)}^{2}}-{{(\sqrt{3})}^{2}}\]
    \[\Rightarrow \]   \[{{x}^{2}}-4x+4-3\]
    \[\Rightarrow \]   \[{{x}^{2}}-4x+1=g(x)\] (say)
    Now, g (x) will be a factor of p(x) so g (x) will be divisible by p (x)
    For other zeroes,
                 \[2{{x}^{2}}-x-1=0\]
               \[2{{x}^{2}}-2x+x-1=0\]
    or          \[2x(x-1)+1\text{ (}x-1)=0\]
            \[(x-1)(2x+1)=0\]
            \[x-1=0\,\,\,\,2x+1=0\]
               \[x=1,x=\frac{-1}{2}\]
    Zeroes of p(x) are
            \[1,\frac{-1}{2},2+\sqrt{3}\] and \[2-\sqrt{3}\].


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