Question

How do you solve by completing the square: ${x^2} + 3x - 8 = 0?$

Answer 1

Hint:In completing the square method we try to fit the given equation into either of the algebraic identity ${(a + b)^2}\;{\text{or}}\;{(a - b)^2}$ with help of algebraic operations like adding required number to become whole square or subtracting and to maintain the balance of the equation we perform the inverse algebraic operation too (inverse of addition is subtraction and vice versa). After getting the form
${(a + b)^2}\;{\text{or}}\;{(a - b)^2}$, we will make the rest term a perfect square or either make it square of some number then use the following algebraic identity :
${x^2} - {y^2} = (x + y)(x - y)$

Complete step by step solution:
In order to solve the given equation ${x^2} + 3x - 8 = 0$ by completing square method, we need to make a complete square using terms with degree two and one as follows
$ \Rightarrow {x^2} + 3x - 8 = 0$
Now, we will make the expression at left hand side a complete square (except constant) in form of either ${(a + b)^2}\;{\text{or}}\;{(a - b)^2}$ and to get the second term we will take $2x$ common from the term with degree one as follows
$ \Rightarrow {x^2} + 2x \times \dfrac{3}{2} - 8 = 0$
So we get the second term equals $\dfrac{3}{2}$
Now adding and subtracting square of $\dfrac{3}{2}$, we will get
\[
\Rightarrow {x^2} + 2 \times x \times \dfrac{3}{2} + {\left( {\dfrac{3}{2}} \right)^2} - {\left(
{\dfrac{3}{2}} \right)^2} - 8 = 0 \\
\Rightarrow {x^2} + 2 \times x \times \dfrac{3}{2} + {\left( {\dfrac{3}{2}} \right)^2} - \dfrac{9}{4} - 8
= 0 \\
\Rightarrow {x^2} + 2 \times x \times \dfrac{3}{2} + {\left( {\dfrac{3}{2}} \right)^2} - \dfrac{{9 + 32}}{4} = 0 \\
\Rightarrow {x^2} + 2 \times x \times \dfrac{3}{2} + {\left( {\dfrac{3}{2}} \right)^2} - \dfrac{{41}}{4}
= 0 \\
\]
We can see that expression at the left hand side seems similar to the equation of square of sum of two numbers, so we write it as follows
$ \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} - \dfrac{{41}}{4} = 0$
Now, we can also write it as
$ \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} - {\left( {\sqrt {\dfrac{{41}}{4}} } \right)^2} = 0$
Using the algebraic identity ${x^2} - {y^2} = (x + y)(x - y)$
$ \Rightarrow \left( {\left( {x + \dfrac{3}{2}} \right) + \sqrt {\dfrac{{41}}{4}} } \right)\left( {\left( {x +
\dfrac{3}{2}} \right) - \sqrt {\dfrac{{41}}{4}} } \right) = 0$
Comparing both factors with zero separately,
$
\Rightarrow \left( {x + \dfrac{3}{2}} \right) + \sqrt {\dfrac{{41}}{4}} = 0\;{\text{and}}\;\left( {x +
\dfrac{3}{2}} \right) - \sqrt {\dfrac{{41}}{4}} = 0 \\
\Rightarrow x = - \dfrac{3}{2} - \sqrt {\dfrac{{41}}{4}} \;{\text{and}}\;x = \sqrt {\dfrac{{41}}{4}} -
\dfrac{3}{2} \\
\Rightarrow x = \dfrac{{ - 3 - \sqrt {41} }}{2}\;{\text{and}}\;x = \dfrac{{\sqrt {41} - 3}}{2} \\
$
$\therefore x = \dfrac{{ - 3 - \sqrt {41} }}{2}\;{\text{and}}\;x = \dfrac{{\sqrt {41} - 3}}{2}$ are required solutions
Formula used:
1. ${(a + b)^2} = {a^2} + 2ab + {b^2}$
2. ${x^2} - {y^2} = (x + y)(x - y)$

Note: In case when roots come to be imaginary, you can use the algebraic identity ${x^2} - {y^2} = (x + y)(x - y)$ with use of iota (i), because when imaginary root comes then you will get positive constants in the equation that will make squares with positive sign, but we can see in the identity we need a negative sign and iota works for that negative sign.