# Solving Quadratic Equations by Completing Square

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

In order to solve the quadratic equation by completing square method, we have the following steps.

1. Write equation in standard form.
2. Shift constant term on RHS.
3. Make 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 method of completing square.

Solution:

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

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

Make the coefficient of ${x^2}$as $1$ 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 ${\left[ {\frac{1}{2}\left( {\frac{7}{5}} \right)} \right]^2} = {\left( {\frac{7}{{10}}} \right)^2}$on 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 square root on 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 method of completing square.

Solution:

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

Multiplied by $- 1$ to make coefficient of ${x^2}$ as positive
${x^2} - 10x + 16 = 0$

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

Now adding ${\left( { - \frac{{10}}{2}} \right)^2} = {\left( 5 \right)^2}$on 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 square root on both sides, we get
$\sqrt {{{\left( {x - 5} \right)}^2}} = \pm \sqrt 9$
$\therefore {\text{ }}x - 5 = \pm 3$

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