Question

Solve the quadratic equation \[2{x^2} + x - 4 = 0\] by completing the square.

Answer 1

Hint: Here, we will solve this quadratic equation by completing the square. Firstly we will simplify the equation by making the coefficient of \[{x^2}\] as 1. Then we will add a constant to both sides to make a perfect quadratic equation on the one side of the equation to get the value of \[x\].

Complete step-by-step answer:
The quadratic equation is \[2{x^2} + x - 4 = 0\].
Firstly, we will modify the given equation by making the coefficient of \[{x^2}\] as 1. Therefore dividing the whole equation by 2, we get
\[ \Rightarrow \dfrac{{2{x^2} + x - 4}}{2} = 0\]
\[ \Rightarrow {x^2} + \dfrac{x}{2} - 2 = 0\]
Adding 2 on both sides, we get
\[ \Rightarrow {x^2} + \dfrac{x}{2} = 2\]
Now we will add \[\dfrac{1}{{16}}\] to both the sides of the equation to get a perfect square term on one side of the equation. Therefore, we get
\[ \Rightarrow {x^2} + \dfrac{x}{2} + \dfrac{1}{{16}} = 2 + \dfrac{1}{{16}}\]
We can write the above equation as
\[ \Rightarrow {x^2} + 2 \times x \times \dfrac{1}{4} + \dfrac{1}{{{4^2}}} = \dfrac{{\left( {2 \times 16} \right) + 1}}{{16}}\]
By using the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\], we get
\[ \Rightarrow {\left( {x + \dfrac{1}{4}} \right)^2} = \dfrac{{32 + 1}}{{16}} = \dfrac{{33}}{{16}}\]
Now we will take the square root on the right hand side of the equation. Therefore,we get
\[ \Rightarrow x + \dfrac{1}{4} = \pm \sqrt {\dfrac{{33}}{{16}}} \]
Now we will take the term \[\dfrac{1}{4}\] to the other side of the equation, we get
\[ \Rightarrow x = \pm \sqrt {\dfrac{{33}}{{16}}} - \dfrac{1}{4} = \pm \dfrac{{\sqrt {33} }}{4} - \dfrac{1}{4}\]
We can write value of \[x\] as
 \[ \Rightarrow x = \dfrac{{ - 1 + \sqrt {33} }}{4},\dfrac{{ - 1 - \sqrt {33} }}{4}\]
Hence, the quadratic equation is solved by completing the square.

Note: Here the one thing we need to keep in mind is that the coefficient of \[{x^2}\] should be 1. If it is not 1 then divide the entire equation by the coefficient of \[{x^2}\]. Also for completing the square we need to add a constant such that the equation becomes the perfect square. The constant should be equal to the square of the half of the middle term. That is the constant should be equal to \[{\left( {\dfrac{b}{2}} \right)^2}\].
In this question \[b = \dfrac{1}{2}\], so substituting its value in \[{\left( {\dfrac{b}{2}} \right)^2}\], we get
Constant \[ = {\left( {\dfrac{{\dfrac{1}{2}}}{2}} \right)^2} = {\left( {\dfrac{1}{4}} \right)^2}\]
Applying the exponent, we get
\[ \Rightarrow \] Constant \[ = \dfrac{1}{{16}}\]