For AI agents: use /llms.txt for the Nakafa content index.
Completing the square is a method for solving quadratic equations by converting the equation from the form a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 to the form ( x + p ) 2 = q (x + p)^2 = q ( x + p ) 2 = q . This method is particularly useful for quadratic equations that are difficult to factor using regular factorization.
Remember that a perfect square trinomial follows the pattern x 2 + 2 p x + p 2 = ( x + p ) 2 x^2 + 2px + p^2 = (x + p)^2 x 2 + 2 p x + p 2 = ( x + p ) 2 . We use this pattern to transform quadratic equations into a more solvable form.
Not all quadratic equations are easily factored. For example, the equation x 2 + 4 x + 2 = 0 x^2 + 4x + 2 = 0 x 2 + 4 x + 2 = 0 cannot be easily factored using rational numbers because there are no two numbers that multiply to give 2 2 2 and add up to 4 4 4 .
In such cases, the completing the square method becomes an effective choice for finding the roots of the equation.
Here are the steps to solve a quadratic equation a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 using the completing the square method:
Ensure the coefficient of x 2 x^2 x 2 is 1 1 1
If the coefficient a a a of x 2 x^2 x 2 is not 1 1 1 , divide the entire equation by the value of a a a .
Example: For the equation 2 x 2 + 6 x + 3 = 0 2x^2 + 6x + 3 = 0 2 x 2 + 6 x + 3 = 0
Move the Constant Term to the Right Side
Move the constant term to the right side of the equation.
Example: From the equation x 2 + 3 x + 3 2 = 0 x^2 + 3x + \frac{3}{2} = 0 x 2 + 3 x + 2 3 = 0
Add the Square of Half the Coefficient of x to Both Sides
Add ( b 2 ) 2 \left(\frac{b}{2}\right)^2 ( 2 b ) 2 to both sides of the equation. This value is the square of half the coefficient of x x x .
Example: For the equation
Factor the Left Side into a Perfect Square
The left side now has the form x 2 + b x + ( b 2 ) 2 x^2 + bx + \left(\frac{b}{2}\right)^2 x 2 + b x + ( 2 b ) 2 , which can be factored as .
Simplify the Right Side
Perform calculations on the right side to get a simpler form.
Example: For ( x + 3 2 ) 2 = − 3 2 + 9 4 \left(x + \frac{3}{2}\right)^2 = -\frac{3}{2} + \frac{9}{4} ( x + 2 3 ) 2 = − 2 3 + 4 9
Take the Square Root of Both Sides
To eliminate the square, take the square root of both sides.
Example: From the equation ( x + 3 2 ) 2 = 3 4 \left(x + \frac{3}{2}\right)^2 = \frac{3}{4} ( x + 2 3 ) 2 = 4 3
Solve for the Unknown Value
Isolate the variable x x x to find the roots of the equation.
Example: From x + 3 2 = ± 3 2 x + \frac{3}{2} = \pm\frac{\sqrt{3}}{2} x + 2 3 = ± 2 3
Let's solve the equation: x 2 + 5 x + 6 = 0 x^2 + 5x + 6 = 0 x 2 + 5 x + 6 = 0
Step 1 1 1 : The coefficient a = 1 a = 1 a = 1 , so we proceed to the next step.
Step 2 2 2 : Move the constant to the right side.
Step 3 3 3 : Add the square of half the coefficient of x x x to both sides.
Step 4 4 4 : Factor the left side into a perfect square.
Step 5 5 5 : Simplify the right side.
Step 6 6 6 : Take the square root of both sides.
Step 7 7 7 : Solve for the value of x x x .
Therefore, the roots of the equation are x = − 2 x = -2 x = − 2 and x = − 3 x = -3 x = − 3 .
Let's solve the equation: 2 x 2 + 6 x + 3 = 0 2x^2 + 6x + 3 = 0 2 x 2 + 6 x + 3 = 0
Step 1 1 1 : Divide all terms by the coefficient a = 2 a = 2 a = 2
Step 2 2 2 : Move the constant to the right side
Step 3 3 3 : Add the square of half the coefficient of x x x to both sides
Step 4 4 4 : Factor the left side into a perfect square
Step 5 5 5 : Simplify the right side
Step 6 6 6 : Take the square root of both sides
Step 7 7 7 : Solve for the value of x x x
Therefore, the roots of the equation are x = − 3 2 + 3 2 x = -\frac{3}{2} + \frac{\sqrt{3}}{2} x = − 2 3 + 2 3 and x = − 3 2 − 3 2 x = -\frac{3}{2} - \frac{\sqrt{3}}{2} x = − 2 3 − 2 3 .
For equations with coefficient of x 2 x^2 x 2 not equal to 1 1 1 : Always divide the entire equation by the coefficient a a a first. Example: 3 x 2 + 6 x + 2 = 0 3x^2 + 6x + 2 = 0 3 x 2 + 6 x + 2 = 0 becomes x 2 + 2 x + 2 3 = 0 x^2 + 2x + \frac{2}{3} = 0 x 2 + 2 x + 3 2 = 0
Constant to be added : Always add the square of half the coefficient of x x x to both sides. Example: For x 2 + 8 x = 5 x^2 + 8x = 5 x 2 + 8 x = 5 , add ( 8 2 ) 2 = 16 \left(\frac{8}{2}\right)^2 = 16 ( 2 8 ) 2 = 16 to both sides.
Final form : The equation will transform into the form ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} ( x + 2 a b ) 2 = 4 a 2 b 2 − 4 a c . Example: becomes
If b 2 − 4 a c < 0 b^2 - 4ac < 0 b 2 − 4 a c < 0 , then the equation has no real roots.
Concrete example: x 2 + 2 x + 2 = 0 x^2 + 2x + 2 = 0 x 2 + 2 x + 2 = 0
Since no real number has a square of − 1 -1 − 1 , this equation has no real roots.
For equations of the form a x 2 + c = 0 ax^2 + c = 0 a x 2 + c = 0 , we don't need to complete the square.
Concrete example: 3 x 2 − 12 = 0 3x^2 - 12 = 0 3 x 2 − 12 = 0
Therefore, the roots of the equation are x = 2 x = 2 x = 2 and x = − 2 x = -2 x = − 2 .
Solve the following quadratic equations using the completing the square method:
x 2 + 5 x + 6 = 0 x^2 + 5x + 6 = 0 x 2 + 5 x + 6 = 0
2 x 2 + 6 x + 3 = 0 2x^2 + 6x + 3 = 0 2 x 2 + 6 x + 3 = 0
6 x 2 + 2 x + 1 6 = 0 6x^2 + 2x + \frac{1}{6} = 0 6 x 2 + 2 x + 6 1 = 0
x 2 − 12 x − 15 = 0 x^2 - 12x - 15 = 0 x 2 − 12 x − 15 = 0
3 2 x 2 − 8 x − 6 = 0 \frac{3}{2}x^2 - 8x - 6 = 0 2 3 x 2 − 8 x − 6 = 0
x 2 + 5 x + 6 = 0 x^2 + 5x + 6 = 0 x 2 + 5 x + 6 = 0
Move the constant to the right side:
Add the square of half the coefficient of x x x to both sides:
Factor the left side into a perfect square:
Simplify the right side:
Take the square root of both sides:
Solve for the value of x x x :
Therefore, the roots of the equation are x = − 2 x = -2 x = − 2 and x = − 3 x = -3 x = − 3 .
2 x 2 + 6 x + 3 = 0 2x^2 + 6x + 3 = 0 2 x 2 + 6 x + 3 = 0
Divide all terms by the coefficient a = 2 a = 2 a = 2 :
6 x 2 + 2 x + 1 6 = 0 6x^2 + 2x + \frac{1}{6} = 0 6 x 2 + 2 x + 6 1 = 0
Divide all terms by the coefficient a = 6 a = 6 a = 6 :
x 2 − 12 x − 15 = 0 x^2 - 12x - 15 = 0 x 2 − 12 x − 15 = 0
Move the constant to the right side:
3 2 x 2 − 8 x − 6 = 0 \frac{3}{2}x^2 - 8x - 6 = 0 2 3 x 2 − 8 x − 6 = 0
Divide all terms by the coefficient a = 3 2 a = \frac{3}{2} a = 2 3 :
x 2 + 3 x = − 3 2 x^2 + 3x = -\frac{3}{2} x 2 + 3 x = − 2 3 Half of the coefficient of x x x is 3 2 \frac{3}{2} 2 3
The square of this value: ( 3 2 ) 2 = 9 4 \left(\frac{3}{2}\right)^2 = \frac{9}{4} ( 2 3 ) 2 = 4 9
( x + b 2 ) 2 \left(x + \frac{b}{2}\right)^2 ( x + 2 b ) 2 Example: From the equation x 2 + 3 x + 9 4 = − 3 2 + 9 4 x^2 + 3x + \frac{9}{4} = -\frac{3}{2} + \frac{9}{4} x 2 + 3 x + 4 9 = − 2 3 + 4 9
Therefore, the roots of the equation are x = − 3 2 + 3 2 x = -\frac{3}{2} + \frac{\sqrt{3}}{2} x = − 2 3 + 2 3 and x = − 3 2 − 3 2 x = -\frac{3}{2} - \frac{\sqrt{3}}{2} x = − 2 3 − 2 3 .
x 2 + 6 x + 8 = 0 x^2 + 6x + 8 = 0 x 2 + 6 x + 8 = 0 ( x + 3 ) 2 = 1 \left(x + 3\right)^2 = 1 ( x + 3 ) 2 = 1
Move the constant to the right side:
Add the square of half the coefficient of x x x to both sides:
Factor the left side into a perfect square:
Take the square root of both sides:
Solve for the value of x x x :
Therefore, the roots of the equation are x = − 3 2 + 3 2 x = -\frac{3}{2} + \frac{\sqrt{3}}{2} x = − 2 3 + 2 3 and x = − 3 2 − 3 2 x = -\frac{3}{2} - \frac{\sqrt{3}}{2} x = − 2 3 − 2 3 .
Move the constant to the right side:
Add the square of half the coefficient of x x x to both sides:
Factor the left side into a perfect square:
Take the square root of both sides:
Solve for the value of x x x :
Therefore, this equation has one (double) root, which is x = − 1 6 x = -\frac{1}{6} x = − 6 1 .
Add the square of half the coefficient of x x x to both sides:
Factor the left side into a perfect square:
Take the square root of both sides:
Solve for the value of x x x :
Therefore, the roots of the equation are x = 6 + 51 x = 6 + \sqrt{51} x = 6 + 51 and x = 6 − 51 x = 6 - \sqrt{51} x = 6 − 51 .
Move the constant to the right side:
Add the square of half the coefficient of x x x to both sides:
Factor the left side into a perfect square:
Take the square root of both sides:
Solve for the value of x x x :
Therefore, the roots of the equation are x = 6 x = 6 x = 6 and x = − 2 3 x = -\frac{2}{3} x = − 3 2 .