Can anyone teach me how to solve a Quadratic equation by completing the square? Thanks..

1 Answer
Apr 5, 2016

The idea behind completing the square is to add or subtract a constant to obtain the form #(x-h)^2# and then take a square root to be left with a linear equation. Let's do a concrete example first.

Starting from #2x^2-7x-4=0#

Step 1: Divide both sides by #2# to obtain #x^2# as the first term

#x^2-7/2x-2 = 0#

Step 2: Add #2# to both sides to isolate the #x# terms.

#x^2-7/2x = 2#

Step 3: Add a constant to both sides which will allow us to factor the left hand side as #(x-h)^2#. Noting that #(x-h)^2 = x^2-2h+h^2# we have #-2h = -7/2# and thus #h = 7/4#, meaning we add #(7/4)^2 = 49/16# to both sides.

#x^2-7/2+49/16 = 81/16#

Step 4: Factor the left hand side

#(x-7/4)^2 = 81/16#

Step 5: Take the square root of both sides. Remember to account for both positive and negative roots.

#x-7/4 = +-sqrt(81/16) = +-9/4#

Step 6: Solve the remaining linear equation:

#x = 7/4 +- 9/4 = 1/4(7+-9)#

#=> x = 4# or #x = -1/2#

The real trick here is observing in step 3 that the constant we need to add is equal to the square of half of the coefficient of #x#.


Let's see what happens if we apply this to a general quadratic equation.

#ax^2 + bx + c = 0#

#=> x^2 + b/ax + c/a = 0#

#=> x^2 + b/ax = -c/a#

#=> x^2 + b/ax + (b/(2a))^2 = -c/a + (b/(2a))^2 = b^2/(4a^2)-c/a#

#=> (x+b/(2a))^2 = (b^2-4ac)/(4a^2)#

#=> x+b/(2a) = +-sqrt((b^2-4ac)/(4a^2)) = +-sqrt(b^2-4ac)/(2a)#

#=> x = -b/(2a) +- sqrt(b^2-4ac)/(2a)#

#=(-b+-sqrt(b^2-4ac))/(2a)#

And we have just derived the quadratic formula.