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
B
316
C
34
D
34

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 squareMathematics

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