Question

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

A
916
B
316
C
34
D
316

Solution

## The correct option is B $$\dfrac { 3 }{ 16 }$$$$4x^2-\sqrt{3}x-5=0$$$$\Rightarrow (2x)^2-\dfrac{\sqrt{3}}{2}2x-5=0$$$$\Rightarrow (2x)^2-2\dfrac{\sqrt{3}}{4}2x+\left(\dfrac{\sqrt{3}}{4}\right)^2-\left(\dfrac{\sqrt{3}}{4}\right)^2-5=0$$$$\Rightarrow \left(2x-\dfrac{\sqrt{3}}{4}\right)^2-\dfrac{3}{16}-5=0$$      ...{$$\because$$  We know, $$a^2-2ab+b^2=(a-b)^2$$}Hence, we had to add $$\dfrac{3}{16}$$ to complete the square.Maths

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