Question

How do you solve by completing the square, leaving answers in simplest radical form: $ {x^2} + 3x - 2 = 0 $ ?

Answer 1

Hint: We have been given to solve a quadratic equation in one variable. Solving the equation by completing the square means finding the value of $ x $ by converting the equation such that we get a perfect square expression in variable $ x $ on one side of the equation. The quadratic equation is converted in the form of $ {\left( {x + m} \right)^2} + n = 0 $ where $ m $ and $ n $ are constant terms.

Complete step by step solution:
We have to solve the given quadratic equation $ {x^2} + 3x - 2 = 0 $ by completing the square. We have to convert the equation such that we get a perfect square expression in variable $ x $ .
We can write the given equation in the form as,
 $
  {x^2} + 3x - 2 = 0 \\
   \Rightarrow {x^2} + \left( {2 \times \dfrac{3}{2} \times x} \right) - 2 = 0 \;
  $
We can add and subtract $ {\left( {\dfrac{3}{2}} \right)^2} $ in the left side of the equation.
 $
  {x^2} + \left( {2 \times \dfrac{3}{2} \times x} \right) - 2 = 0 \\
   \Rightarrow {x^2} + \left( {2 \times \dfrac{3}{2} \times x} \right) + {\left( {\dfrac{3}{2}} \right)^2} - 2 - {\left( {\dfrac{3}{2}} \right)^2} = 0 \;
  $
We can see that the first three terms are in the form of $ {x^2} + 2ax + {a^2} $ which can be written as $ {\left( {x + a} \right)^2} $ using the formula $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ .
 $
   \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} - 2 - \dfrac{9}{4} = 0 \\
   \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} - \left( {2 + \dfrac{9}{4}} \right) = 0 \\
   \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} - \left( {\dfrac{{8 + 9}}{4}} \right) = 0 \\
   \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} - \dfrac{{17}}{4} = 0 \;
  $
We can see that we got the expression containing the variable $ x $ in the form of a perfect square.
Now to solve this equation we transfer the constant term to the right side and take the square root both sides.
 $
  {\left( {x + \dfrac{3}{2}} \right)^2} - \dfrac{{17}}{4} = 0 \\
   \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} = \dfrac{{17}}{4} \\
   \Rightarrow \left( {x + \dfrac{3}{2}} \right) = \pm \sqrt {\dfrac{{17}}{4}} = \pm \dfrac{{\sqrt {17} }}{2} \\
   \Rightarrow x = - \dfrac{3}{2} \pm \dfrac{{\sqrt {17} }}{2} \;
  $
Now we will try to simplify the result to get the result in the simplest form.
 $ x = - \dfrac{3}{2} \pm \dfrac{{\sqrt {17} }}{2} = \dfrac{{ - 3 \pm \sqrt {17} }}{2} $
We can see that we cannot simplify this further.
Hence, $ x = \dfrac{{ - 3 + \sqrt {17} }}{2}\;\;or\;\;\dfrac{{ - 3 - \sqrt {17} }}{2} $ is the solution for the given equation.
So, the correct answer is “ $ x = \dfrac{{ - 3 + \sqrt {17} }}{2}\;\;or\;\;\dfrac{{ - 3 - \sqrt {17} }}{2} $ ”.

Note: We completed the square in the given equation by transforming the equation such that we got it in the form $ {\left( {x + m} \right)^2} + n = 0 $ . Further we transferred the constant term to the right side and then took square roots on both sides to find the solution. Solution in simplest radical form means that terms cannot be simplified further. Since we are given a quadratic equation, we got two values of $ x $ in the solution.