Completing the Square when a = 1
Where a, b, and c are constants and a ≠ 0. In other words there must be a x^{2} term.
Some examples are:
x^{2} + 3x  3 = 0
4x^{2} + 9 = 0 (Where b = 0)
x^{2} + 5x = 0 (where c = 0)
One way to solve a quadratic equation is by completing the square.
Where r and s are constants.
This discussion will focus on completing the square when a, the x^{2} coefficient, is 1.
Let's solve the following equation by completing the square: x^{2} = 6x  7 

Step 1: Write the equation in the general form ax^{2} + bx + c = 0. 
x^{2}  6x + 5 = 0 
Step 2: Move c, the constant term, to the righthand side of the equation. 
c = 5 x^{2}  6x = 5 
Step 3: Divide b, the xcoefficient, by two and square the result. 
b = 6 $\frac{6}{2}={}{3}\to {r}$ (3)^{2} = 9 
Step 4: Add the result from Step 3 to both sides of the equation. 
x^{2}  6x + 9 = 5 + 9 
Step 5: Rewrite the lefthand side as a perfect square and simplify the righthand side. When rewriting in perfect square format the value in the parentheses is the b, xcoefficient, divided by 2 as found in Step 3. 
(x  3)^{2} = 5 + 9 (x  3)^{2} = 4 
Now that the square has been completed, solve for x. 

Step 6: Take the square root of both sides of the equation. Remember that when taking the square root on the righthand side the answer can be positive or negative. 
x  3 = ± 2 
Step 7: Solve for x. 
x = 3 ± 2 x = 5 or x = 1 
Example 1: x^{2} + 2x  5 = 0 

Step 1: Write the equation in the general form ax^{2} + bx + c = 0. This equation is already in the proper form. 
x^{2} + 2x  5 = 0 
Step 2: Move c, the constant term, to the righthand side of the equation. 
c = 5 x^{2} + 2x = 5 
Step 3: Divide b, the x coefficient, by two and square the result. 
b = 2 $\frac{2}{2}={1}\to {r}$ (1)^{2} = 1 
Step 4: Add the result from Step 3 to both sides of the equation. 
x^{2} + 2x + 1 = 5 + 1 
Step 5: Rewrite the lefthand side as a perfect square and simplify the righthand side. When rewriting in perfect square format the value in the parentheses is the b, xcoefficient, divided by 2 as found in Step 3. 
(x + 1)^{2} = 5 + 1 (x + 1)^{2} = 6 
Now that the square has been completed, solve for x. 

Step 6: Take the square root of both sides of the equation. Remember that when taking the square root on the righthand side the answer can be positive or negative. 
$x+1=\pm \sqrt{6}$ 
Step 7: Solve for x. 
$x=1\pm \sqrt{6}$ 
Example 2:
${x}^{2}=\frac{1}{2}x$


Step 1: Write the equation in the general form ax^{2} + bx + c = 0. 
${x}^{2}\frac{1}{2}x=0$ 
Step 2: Move c, the constant term, to the righthand side of the equation. There is no c in this equation. 
${x}^{2}\frac{1}{2}x=0$ 
Step 3: Divide b, the xcoefficient, by two and square the result. 
${b}=\frac{1}{2}$ $\frac{1}{2}\xf72={}\frac{{1}}{{4}}\to {r}$ ${\left(\frac{1}{4}\right)}^{2}=\frac{{1}}{{16}}$ 
Step 4: Add the result from Step 3 to both sides of the equation. 
${x}^{2}\frac{1}{2}x+\frac{{1}}{{16}}=0+\frac{{1}}{{16}}$ 
Step 5: Rewrite the lefthand side as a perfect square and simplify the righthand side. When rewriting in perfect square format the value in the parentheses is the b, xcoefficient, divided by 2 as found in Step 3. 
${\left(x{}\frac{{1}}{{4}}\right)}^{2}=0+\frac{1}{16}$ ${\left(x{}\frac{{1}}{{4}}\right)}^{2}=\frac{{1}}{{16}}$ 
Now that the square has been completed, solve for x. 

Step 6: Take the square root of both sides of the equation. Remember that when taking the square root on the righthand side the answer can be positive or negative. 
$x\frac{1}{4}=\pm \sqrt{\frac{1}{16}}$ 
Step 7: Solve for x. 
$x=\frac{1}{4}\pm \frac{1}{4}$ $x=\frac{1}{2}orx=0$ 
PART II of Completing the Square addresses the case when a ≠ 1.
Related Links: Math algebra Completing the Square when a ≠ 1 Factoring Quadratic Equations when a equals 1 Algebra Topics 
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