Question

How do you solve by completing the square: \[{x^2} + 3x - 4 = 0\]?

Answer 1

Hint: We will first write the equation \[{x^2} + 3x - 4 = 0\] in the form \[{x^2} + 2 \times a \times x + {a^2} - {a^2} + c = 0\] so that we may use it in the form ${(x + a)^2} = {x^2} + {a^2} + 2ax$ and thus we have completed the square.

Complete step by step solution:
We are given that we are required to solve the equation \[{x^2} + 3x - 4 = 0\] by completing the square.
We can write the given equation \[{x^2} + 3x - 4 = 0\] as \[{\left( x \right)^2} + 2 \times \left( {\dfrac{3}{2}} \right) \times x + {\left( {\dfrac{3}{2}} \right)^2} - {\left( {\dfrac{3}{2}} \right)^2} - 4 = 0\].
Simplifying the above equation, we can also write this as: \[{x^2} + 2 \times \left( {\dfrac{3}{2}} \right) \times x + {\left( {\dfrac{3}{2}} \right)^2} - \dfrac{9}{4} - 4 = 0\]. ……..(1)
Now, we know that we have a formula given by \[{(a + b)^2} = {a^2} + 2 \times a \times b + {b^2}\].
Replacing a by $x$ and b by $\dfrac{3}{2}$ in the above mentioned formula, we will then obtain the following equation:-
\[ \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} = {x^2} + 2 \times x \times \left( {\dfrac{3}{2}} \right) + {\left( {\dfrac{3}{2}} \right)^2}\]
Putting this in equation number (1), we will then obtain the given equation \[{x^2} + 3x - 4 = 0\] can be written as follows:-
\[ \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} - \dfrac{9}{4} - 4 = 0\]
Taking the constants on the right hand side, we will then obtain the following equation:-
\[ \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} = \dfrac{9}{4} + 4\]
Simplifying the calculations by taking least common multiple on the right hand side, we will then obtain the following equation:-
\[ \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} = \dfrac{{25}}{4}\]
Taking square – root on both the sides of the above equation, we will then obtain the following equation:-
\[ \Rightarrow x + \dfrac{3}{2} = \pm \dfrac{5}{2}\]
Taking \[\dfrac{3}{2}\] from addition in the left hand side to subtraction in the right hand side, we will then obtain the following equation:-
\[ \Rightarrow x = - \dfrac{3}{2} \pm \dfrac{5}{2}\]

Therefore, we have \[x = 1, - 4\].

Note:
The students must note that there is an alternate way to solve it, if not mentioned that we need to solve it by ‘completing the square’.
Alternate Way:
We are given that we are required to solve the equation \[{x^2} + 3x - 4 = 0\]
We can write this as: \[{x^2} - x + 4x - 4 = 0\]
Taking $x$ common from first two terms and 4 common from the last two terms in the above equation, we will then obtain the following equation:-
\[ \Rightarrow x\left( {x - 1} \right) + 4\left( {x - 1} \right) = 0\]
Taking \[\left( {x - 1} \right)\] common from the equation, we will then obtain the following equation:
\[ \Rightarrow \left( {x - 1} \right)\left( {x + 4} \right) = 0\]
Thus, we have the possible values of x as 1 and – 4.