Question

How do you solve $ 4{{x}^{2}}+2x-1=0 $ by completing the square?

Answer 1

Hint:
 In this question, we need to find the value of x from the equation $ 4{{x}^{2}}+2x-1=0 $ by completing the square. For this, we will first make coefficient of $ {{x}^{2}} $ as 1 by dividing equation by 4. We will add and subtract a term y in the left side. y for an equation $ {{x}^{2}}+bx+c $ is given by $ {{\left( \dfrac{b}{2} \right)}^{2}} $ . After that, we will use positive y to form a whole square of the form $ {{\left( a+b \right)}^{2}} $ and take the left out constant term on the other side, at last, taking square root on both sides and solving for x will give us find value of x.

Complete step by step answer:
Here we have the equation as $ 4{{x}^{2}}+2x-1=0 $ . We need to solve it by completing the square. For this, let us add and subtract a certain element y in the left side of the equation.
We know that, for any equation of the form $ {{x}^{2}}+bx+c $ we add and subtract $ {{\left( \dfrac{b}{2} \right)}^{2}} $ element to complete the square. But here we have a coefficient of $ {{x}^{2}} $ . So we need to remove it first to compare it with $ {{x}^{2}}+bx+c $ .
Dividing the whole equation by 4 we get, $ \dfrac{4{{x}^{2}}+2x-1}{4}=\dfrac{0}{4}\Rightarrow {{x}^{2}}+\dfrac{1}{2}x-\dfrac{1}{4}=0 $ .
Now comparing with $ {{x}^{2}}+bx+c $ . We have $ b=\dfrac{1}{2} $ .
Therefore $ y={{\left( \dfrac{\dfrac{1}{2}}{2} \right)}^{2}}={{\left( \dfrac{1}{4} \right)}^{2}} $ .
Adding and subtracting y we get, $ {{x}^{2}}+\dfrac{1}{2}x-\dfrac{1}{4}+{{\left( \dfrac{1}{4} \right)}^{2}}-{{\left( \dfrac{1}{4} \right)}^{2}}=0 $ .
Rearranging it we get $ {{x}^{2}}+\dfrac{1}{2}x+{{\left( \dfrac{1}{4} \right)}^{2}}=\dfrac{1}{4}+{{\left( \dfrac{1}{4} \right)}^{2}} $ .
We know $ \dfrac{1}{2} $ can be written as $ 2\cdot \dfrac{1}{4} $ so we have,
 $ {{x}^{2}}+2\cdot \dfrac{1}{4}x+{{\left( \dfrac{1}{4} \right)}^{2}}=\dfrac{1}{4}+\dfrac{1}{16} $ .
As we can see, the left hand side of the equation is of the form $ {{a}^{2}}+2ab+{{b}^{2}} $ . So we can write it as $ {{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}} $ . Hence we have a = x, $ b=\dfrac{1}{4} $ for above equation we get, $ {{\left( x+\dfrac{1}{4} \right)}^{2}}=\dfrac{1}{4}+\dfrac{1}{16} $ .
Taking LCM of 16 in the right hand side of the equation we get $ {{\left( x+\dfrac{1}{4} \right)}^{2}}=\dfrac{4+1}{16}\Rightarrow {{\left( x+\dfrac{1}{4} \right)}^{2}}=\dfrac{5}{16} $ .
Now let us take square root on both sides of the equation we get $ {{\left( {{\left( x+\dfrac{1}{4} \right)}^{2}} \right)}^{\dfrac{1}{2}}}={{\left( \dfrac{5}{16} \right)}^{\dfrac{1}{2}}}\Rightarrow x+\dfrac{1}{4}=\pm \dfrac{\sqrt{5}}{4} $ because $ {{4}^{2}}=16 $ .
Therefore we can say that $ x+\dfrac{1}{4}=\dfrac{\sqrt{5}}{4}\text{ and }x+\dfrac{1}{4}=-\dfrac{\sqrt{5}}{4} $ .
Taking $ \dfrac{1}{4} $ to the other side we get,
 $ x=\dfrac{\sqrt{5}}{4}-\dfrac{1}{4}\text{ and }x=-\dfrac{\sqrt{5}}{4}-\dfrac{1}{4}\Rightarrow x=\dfrac{\sqrt{5}-1}{4}\text{ and }x=-\dfrac{\sqrt{5}-1}{4} $ .
Hence we have obtained the value of x as $ \dfrac{\sqrt{5}-1}{4},\dfrac{-\sqrt{5}-1}{4} $ .

Note:
 Students should not forget to make the coefficient of $ {{x}^{2}} $ as 1 before applying the completing the square method. Students can check their answer by putting the formed value in the original equation to see if the left side is equal to the right side.