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Question

Which constant should be added and subtracted to solve the quadratic equation $$4{ x }^{ 2 }-\sqrt { 3 } x-5=0$$ by the method of completing the square?


A
910
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B
316
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C
34
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D
34
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Solution

The correct option is A $$\displaystyle\frac{3}{16}$$
We have,
$$4{ x }^{ 2 } - \sqrt { 3 } x - 5 = 0$$
Divide whole equation by $$4$$,
$$\therefore   { x }^{ 2 } - \displaystyle\frac { \sqrt { 3 }  }{ 4 } x - \displaystyle\frac { 5 }{ 4 } = 0$$
$$\therefore { \left( x -\displaystyle\frac { \sqrt { 3 }  }{ 8 }  \right)  }^{ 2 } - \displaystyle\frac { 5 }{ 4 } = \displaystyle\frac { 3 }{ 64 }$$
$$\Rightarrow 4{ \left( x -\displaystyle\frac { \sqrt { 3 }  }{ 8 }  \right)  }^{ 2 } - 5 - \displaystyle\frac { 3 }{ 16 } = 0$$
$$\therefore \displaystyle \frac { 3 }{ 16 }$$ must be added to make $$4{ x }^{ 2 } - \sqrt { 3 } x - 5 = 0$$ a complete square

Mathematics

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