Solving Quadratic Equations by Completing Square

Sometimes factorization of a given quadratic equation is not possible. When coefficients in the quadratic equation are large numbers, then it may be difficult to factorize. In this case, we can use the completing square method.

In order to solve the quadratic equation with the completing square method, we have the following steps:

  1. Write the equation in standard form.
  2. Shift the constant term on RHS.
  3. Make the coefficient of {x^2} as one.
  4. Add on both sides {\left( {\frac{1}{2}{\text{ Coefficient of}}x} \right)^2}
  5. Finally, simplify the equation to get the required roots.

 

Example:

Solve the equation by the completing square method.

5{x^2} + 7x + 2 = 0

Solution:

Write the equation in standard form 5{x^2} + 7x + 2 = 0

Shifting the constant term on RHS, we get
5{x^2} + 7x = - 2

Make the coefficient of {x^2}as 1 and divide the equation by 5
\frac{{5{x^2}}}{5} + \frac{{7x}}{5} = - \frac{2}{5}
{x^2} + \frac{7}{5}x = - \frac{2}{5}

Now adding {\left[ {\frac{1}{2}\left( {\frac{7}{5}} \right)} \right]^2} = {\left( {\frac{7}{{10}}} \right)^2} to both sides, we get
{x^2} + \frac{7}{5}x + {\left( {\frac{7}{{10}}} \right)^2} = - \frac{2}{5} + {\left( {\frac{7}{{10}}} \right)^2}
{\left( {x + \frac{7}{{10}}} \right)^2} = - \frac{2}{5} + \frac{{49}}{{100}}
{\left( {x + \frac{7}{{10}}} \right)^2} = \frac{{ - 40 + 49}}{{100}} = \frac{9}{{100}}

Taking the square root of both sides, we get
\sqrt {{{\left( {x + \frac{7}{{10}}} \right)}^2}} = \pm \sqrt {\frac{9}{{100}}}
\therefore {\text{ }}x + \frac{7}{{10}} = \pm \frac{3}{{10}}

Either
x + \frac{7}{{10}} = \frac{3}{{10}}    or    x + \frac{7}{{10}} = - \frac{3}{{10}}
x = \frac{3}{{10}} - \frac{7}{{10}}         x = - \frac{3}{{10}} - \frac{7}{{10}}
x = \frac{{3 - 7}}{{10}}               x = \frac{{ - 3 - 7}}{{10}}
x = - \frac{2}{5}                  x = - 1

Example:

Solve the equation by the completing square method.

- {x^2} - 16 = 0

Solution:

Write the equation in standard form as  - {x^2} + 10x - 16 = 0

Multiply by  - 1 to make the coefficient of {x^2} positive
{x^2} - 10x + 16 = 0

Shifting the constant term on RHS, we get
{x^2} - 10x = - 16

Now adding {\left( { - \frac{{10}}{2}} \right)^2} = {\left( 5 \right)^2} to both sides, we get
{x^2} - 10x + {\left( 5 \right)^2} = - 16 + {\left( 5 \right)^2}
{\left( {x - 5} \right)^2} = - 16 + 25
{\left( {x - 5} \right)^2} = 9

Taking the square root of both sides, we get
\sqrt {{{\left( {x - 5} \right)}^2}} = \pm \sqrt 9
\therefore {\text{ }}x - 5 = \pm 3

Either x - 5 = 3    or   x - 5 = - 3
x = 5 + 3         x = 5 - 3
x = 8        x = 2