Question

How do you solve $ {x^2} + 10x - 1 = 0 $ by completing the square?

Answer 1

Hint: First of all split the factor out the second term in three factors in which one will be two and another will be the variable present in the equation and whatever the third factor remains will become the second term for completing the square. Now add and subtract the third term’s square in order to make a perfect square and then after making the perfect square, solve further to get the required solution.

Complete step by step solution:
In order to solve the given equation by completing the square, we will first split the second term or the term of degree one as follows
 $
   \Rightarrow {x^2} + 10x - 1 = 0 \\
   \Rightarrow {x^2} + 2 \times x \times 5 - 1 = 0 \;
  $
Now, we will add and subtract the equation with $ {5^2} $ , we will get
 $
   \Rightarrow {x^2} + 2 \times x \times 5 + {5^2} - {5^2} - 1 = 0 \\
   \Rightarrow {x^2} + 2 \times x \times 5 + {5^2} - 25 - 1 = 0 \\
   \Rightarrow {x^2} + 2 \times x \times 5 + {5^2} - 26 = 0 \\
   \Rightarrow {x^2} + 2 \times x \times 5 + {5^2} = 26 \;
  $
Now, we can see that the equation is seems to be the algebraic equation of square of addition of two numbers, so we can further write the equation with help of algebraic equation as follows
 $ \Rightarrow {\left( {x + 5} \right)^2} = 26 $
Taking square root both sides of the equation, we will get
 $
   \Rightarrow \left( {x + 5} \right) = \pm \sqrt {26} \\
   \Rightarrow x = \pm \sqrt {26} - 5 \;
  $
Therefore $ x = \pm \sqrt {26} - 5 $ is the required solution of the given quadratic equation.
So, the correct answer is “ $ x = \pm \sqrt {26} - 5 $ ”.

Note: Quadratic equation is the equation with degree two as in this question the given equation was a quadratic equation because it also has a degree of two. Also you can further solve the root part of the solution with help of a calculator or either by yourself (By taking approximate value) to write the answer in decimal number. The equation has a degree of two that’s why we get two solutions.