Question

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

Answer 1

In this question we have given a quadratic equation in one variable for which we need to find a solution of term $x$ using the completing the square method. As we know the property of the square of two numbers as: ${a^2} + 2ab + {b^2} = {(a + b)^2}$ using this and doing some simplification we get the required answer.

Complete step by step answer: So in order to convert the given $4{x^2} - x$ in form ${a^2} + 2ab$ we can have:
$\Rightarrow 4{x^2} - x{\text{ }} = {(2x)^2} + 2 \times 2x \times \left( { - \dfrac{1}{4}} \right)$
An expression like $4{x^2} - x$ may be thought of as an "incomplete" square. To make it complete, we would need to add ${\left( { - \dfrac{1}{4}} \right)^2}$to the expression.
This means we should add ${\left( { - \dfrac{1}{4}} \right)^2}$on both sides for completing the square.
Of course you can't just add a number to an expression without changing the value of the expression, so if we want to keep the same value we'll have to make up for adding ${\left( { - \dfrac{1}{4}} \right)^2}$
We can do this by also subtracting${\left( { - \dfrac{1}{4}} \right)^2}$.
That doesn't change the value of$4{x^2} - x$, but it does change the way it's written.
Hence, $4{x^2} - x{\text{ }} = {\text{ }}0$ can be written as:
$\Rightarrow {(2x)^2} + 2 \times 2x \times \left( { - \dfrac{1}{4}} \right) + {\left( { - \dfrac{1}{4}} \right)^2} - {\left( { - \dfrac{1}{4}} \right)^2} = 0$
On further simplification it would seem as shown:
$\Rightarrow {\left( {2x - \dfrac{1}{4}} \right)^2} - {\left( { - \dfrac{1}{4}} \right)^2} = 0\;$
On using the formula we get
$\; \Rightarrow \left( {2x - \dfrac{1}{4} - \dfrac{1}{4}} \right)\left( {2x - \dfrac{1}{4} + \dfrac{1}{4}} \right) = 0$
Solving an equation by completing the square:
$\Rightarrow \left( {2x - \dfrac{1}{2}} \right)\left( {2x} \right) = 0\;$
Each of the following equations is equivalent (has exactly the same solutions) as the lines before it
$\Rightarrow 2\left( {x - \dfrac{1}{4}} \right) \times 2x = 0\;$
And dividing by 4 and on rearranging the terms we have;
$\Rightarrow x\left( {x - \dfrac{1}{4}} \right) = 0$
i.e. $x = 0\;or\;x = \dfrac{1}{4}$
The solution set to the first equation is: \left\\{ {0,\dfrac{1}{4}} \right\\}
As long as we are comfortable calculating square roots, we can now solve any quadratic equation.
Completing the square is also useful for getting the equation of a circle, ellipse or other conic section into standard form.

Note:
Since we know that factoring is usually faster and one must always try factoring first; although factoring is not always easy, so if we get this kind of case, then we must opt for completing the square method first and then the quadratic formula onwards.
Alternative Quadratic formula:
To find the roots of a quadratic equation in the form $a{x^2}\; + \;bx\; + \;c\; = {\text{ }}0$, follow these steps:
(i) If a does not equal 1, divide each side by a (so that the coefficient of the ${x^2}$ is 1).
${x^2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0$
(ii) Rewrite the equation with the constant term on the right side.
${x^2} + \dfrac{b}{a}x = - \dfrac{c}{a}$
(iii) Complete the square by adding the square of one-half of the coefficient of $x$ (this is the square of$\dfrac{b}{{2a}}$) to both sides.
${x^2} + \dfrac{b}{a}x + \dfrac{{{b^2}}}{{4{a^2}}} = \dfrac{{{b^2}}}{{4{a^2}}} - \dfrac{c}{a}$
(iv) Write the left side as a perfect square and simplify the right side.
${\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{{{b^2} - 4ac}}{{4{a^2}}}$
(v) Equate and solve.
$\left( {x + \dfrac{b}{{2a}}} \right) = \pm \sqrt {\dfrac{{{b^2} - 4ac}}{{4{a^2}}}}$
$x = - \dfrac{b}{{2a}} \pm \sqrt {\dfrac{{{b^2} - 4ac}}{{4{a^2}}}}$
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$