Question

How do you solve the quadratic equation by completing the square $4{x^2} + 9 = 12x$ ?

Answer 1

In this question, we should solve the quadratic equation using complete square, first transform the equation such that the constant term is on the right side, and divide both sides with the coefficient of ${x^2}$ term i.e., 4, now add the square of the half of the coefficient to both sides, here it is ${\left( {\dfrac{{12}}{8}} \right)^2}$,now factor the square of the binomial on the left side, and take the square root on both sides, and solve for required $x$.

Complete step-by-step solution:
Completing the Square is a method used to solve a quadratic equation by changing the form of the equation so that the left side is a perfect square trinomial.
To solve $a{x^2} + bx + c = 0$ by completing the square:
1. Transform the equation so that the constant term,$c$, is alone on the right side.
2. If $a$ the leading coefficient (the coefficient of the ${x^2}$ term), is not equal to 1, divide both sides by $a$
3. Add the square of half the coefficient of the term, ${\left( {\dfrac{b}{{2a}}} \right)^2}$ to both sides of the equation.
4. Factor the left side as the square of a binomial.
5. Take the square root of both sides. (Remember: ${\left( {x + q} \right)^2} = r$ is equivalent to $x + q = \sqrt r$.)
6. Solve for $x$.
Now given quadratic equation, $4{x^2} + 9 = 12x$
Now transforming the equation we get,
$\Rightarrow 4{x^2} - 12x = - 9$,
Now by dividing the term by 4 we get,
\Rightarrow $$$${x^2} - \dfrac{{12}}{4}x = \dfrac{{ - 9}}{4},
Now comparing this to ${a^2} + 2ab + {b^2}$ format, we do the following re-grouping, we further add and subtract ${\left( {\dfrac{{12}}{8}} \right)^2}$ from both the sides we get,
$\Rightarrow {x^2} - 2\left( {\dfrac{{12}}{8}x} \right) + {\left( {\dfrac{{12}}{8}} \right)^2} = \dfrac{{ - 9}}{4} + {\left( {\dfrac{{12}}{8}} \right)^2}$
Now we can see that the half of the expression represents a perfect square, we get,
${\left( {x - \dfrac{{12}}{8}} \right)^2} = - \dfrac{9}{4} + {\left( {\dfrac{3}{2}} \right)^2}$,
Now simplifying we get,
${\left( {x - \dfrac{{12}}{8}} \right)^2} = - \dfrac{9}{4} + \dfrac{9}{4}$,
Now again simplifying we get,
\Rightarrow $$$${\left( {x - \dfrac{{12}}{8}} \right)^2} = 0,
Now taking out the square we get,
$\left( {x - \dfrac{{12}}{8}} \right) = 0$,
Now taking the constant term to right hand side we get,
$x = \dfrac{{12}}{8}$,
Now taking the common term, we get,
$x = \dfrac{3}{2}$,
The correct value for $x$ is $x = \dfrac{3}{2}$.
The value of $x$ when the quadratic equation is solved by completely the square is 0.

Note: In these type of questions, we can solve by using quadratic formula i.e., $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, but we should keep in mind that we can also solve the equation using completely the square, and we can cross check the values of $x$ by using the above formula. Also we should always convert the coefficient of ${x^2} = 1$, to easily solve the equation by this method.