Question

How do you solve the equation $4{x^2} + 9x - 1 = 0$ by completing the square?

Answer 1

Hint: Completing the square is the method which represents the quadratic equation as the combination of the quadrilateral used to form the square and it is the basis of the method discovers the special value which when added to both the sides of the quadratic which creates the perfect square trinomial. Here we will take the given expression and check for the perfect square or the value to be added.

Complete step-by-step solution:
Take the given expression: $4{x^2} + 9x - 1 = 0$
Take out common terms from all the terms in the above equation.
$4\left( {{x^2} + \dfrac{9}{4}x - \dfrac{1}{4}} \right) = 0$ ….. (A)
Now to create the trinomial square on the left hand side of the equation find the value which is equal to the square of half of the “b”
${\left( {\dfrac{b}{2}} \right)^2} = {\left( {\dfrac{9}{{4 \times 2}}} \right)^2}$
Simplify the above equation.
${\left( {\dfrac{b}{2}} \right)^2} = {\left( {\dfrac{9}{8}} \right)^2}$
Add and subtract ${\left( {\dfrac{9}{8}} \right)^2}$in the equation (A)
$4\left( {{x^2} + \dfrac{9}{4}x - \dfrac{1}{4} + {{\left( {\dfrac{9}{8}} \right)}^2} - {{\left( {\dfrac{9}{8}} \right)}^2}} \right) = 0$
The above equation can be rearranged as –
$4\left( {{{\underline {{x^2} + \dfrac{9}{4}x + \left( {\dfrac{9}{8}} \right)} }^2} - \dfrac{1}{4} - {{\left( {\dfrac{9}{8}} \right)}^2}} \right) = 0$
Place the whole square formula for the first three terms and simplify last two terms using LCM (Least common multiple)
The above equation can be written in the simplified form as using the perfect trinomial square which can be framed as ${(x + a)^2}$
$4\left( {{{\underline {\left( {x + \dfrac{9}{8}} \right)} }^2} - \left( {\dfrac{{97}}{{64}}} \right)} \right) = 0$
Now, split the multiple to both the terms inside the bracket.
${\underline {4\left( {x + \dfrac{9}{8}} \right)} ^2} - 4\left( {\dfrac{{97}}{{64}}} \right) = 0$
Common multiple from the numerator and the denominator cancel each other.
${\underline {4\left( {x + \dfrac{9}{8}} \right)} ^2} - \left( {\dfrac{{97}}{{16}}} \right) = 0$
Make the required term as the subject, move other terms on the opposite side.
${\underline {4\left( {x + \dfrac{9}{8}} \right)} ^2} = \left( {\dfrac{{97}}{{16}}} \right)$
Term multiplicative on one side if moved to the opposite side, then it goes to the denominator.
${\underline { \Rightarrow \left( {x + \dfrac{9}{8}} \right)} ^2} = \left( {\dfrac{{97}}{{16 \times 4}}} \right)$
Take the square root on both the sides of the equation.
$ \Rightarrow \sqrt {{{\underline {\left( {x + \dfrac{9}{8}} \right)} }^2}} = \sqrt {\left( {\dfrac{{97}}{{64}}} \right)} $
Square and square root cancel each other.
$ \Rightarrow \left( {x + \dfrac{9}{8}} \right) = \pm \sqrt {\left( {\dfrac{{97}}{{64}}} \right)} $
Make “x” subject
$ \Rightarrow x = \pm \sqrt {\left( {\dfrac{{97}}{{64}}} \right)} - \dfrac{9}{8}$
This is the required solution.

Note: Always remember that the square root of positive term can be positive or negative term while the square of positive or negative term always gives you the positive term. Be good in multiples and squares and square root concepts.