Question

How do you solve by completing the square: ${x^2} + 6x - 5 = 0?$

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 algebraic operations to both sides simultaneously. After getting the form ${(a + b)^2}\;{\text{or}}\;{(a - b)^2}$ we send rest of terms to right hand side of the equation and take square root both sides and then solve for the required variable.

Complete step by step solution:
In order to solve the given equation ${x^2} + 6x - 5 = 0$ by completing square method, we will first send the constant term to right hand side,
$
\Rightarrow {x^2} + 6x - 5 = 0 \\
\Rightarrow {x^2} + 6x = 5 \\
$
Now, we will make the expression at left hand side a complete square 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} + 2 \times x \times 3 = 5$
So we get the second term equals $3$
Now adding both sides square of $3$, we will get
$
\Rightarrow {x^2} + 2 \times x \times 3 + {3^2} = 5 + {3^2} \\
\Rightarrow {x^2} + 2 \times x \times 3 + {3^2} = 5 + 9 \\
\Rightarrow {x^2} + 2 \times x \times 3 + {3^2} = 14 \\
$
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 + 3} \right)^2} = 14$
Now taking square root both sides we will get
\[
\Rightarrow \sqrt {{{\left( {x + 3} \right)}^2}} = \sqrt {14} \\
\Rightarrow \left( {x + 3} \right) = \pm \sqrt {14} \\
\]
Solving for $x$ by taking both cases (positive and negative) separately we will get,
For positive sign,
\[
\Rightarrow \left( {x + 3} \right) = \sqrt {14} \\
\Rightarrow x = \sqrt {14} - 3 \\
\]
And for negative sign,
\[
\Rightarrow \left( {x + 3} \right) = - \sqrt {14} \\
\Rightarrow x = - \sqrt {14} - 3 \\
\Rightarrow x = - \left( {\sqrt {14} + 3} \right) \\
\]
$\therefore x = - \left( {\sqrt {14} + 3} \right)\;and\;\sqrt {14} - 3$ is the required solution.

Note: When you are using completing the square method in an equation which has its coefficient of ${x^2}$ other than one, then divide the whole equation with that coefficient in order to get the coefficient of ${x^2}$ equals one, we do this because it’s easy to make it perfect square rather than other numbers.