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.
PART I of this topic focused on completing the square when a, the x^{2}coefficient, is 1. This part, PART II, will focus on completing the square when a, the x^{2}coefficient, is not 1.
Let's solve the following equation by completing the square:
Step 1: Write the equation in the general form ax^{2} + bx + c = 0. This equation is already in the proper form where a = 2 and c = 5. 
2x^{2} + 8x  5 = 0 
Step 2: Move c, the constant term, to the righthand side of the equation. 
c = 5 2x^{2} + 8x = 5 
Step 3: Factor out a from the lefthand side. This changes the value of the xcoefficient. 
a = 2 2(x^{2} + 4x) = 5 
Step 4: Complete the square of the expression in parentheses on the lefthand side of the equation. The expression is x^{2} + 4x. Divide the xcoefficient by two and square the result. 
x^{2} + 4x xcoefficient = 4 $\frac{4}{2}={2}\to {r}$ (2)^{2} = 4 
Step 5: Add the result from Step 4 to the parenthetical expression on the lefthand side. Then add a x result to the righthand side. To keep the equation true what is done to one side must also be done to the other. When adding the result to the parenthetical expression on the lefthand side the total value added is a x result. So this value must also be added to the righthand side. 
2(x^{2} + 4x + 4) = 5 + 2(4) 
Step 6: 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 xcoefficient of the parenthetical expression divided by 2 as found in Step 4. 
2(x + 2)^{2} = 13 
Now that the square has been completed, solve for x. 

Step 7: Divide both sides by a. 
${\left(x+2\right)}^{2}=\frac{13}{{2}}$ 
Step 8: 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. 
$x+2=\pm \sqrt{\frac{13}{2}}$ 
Step 9: Solve for x. 
$x=2\pm \sqrt{\frac{13}{2}}$ 
Step 1: Write the equation in the general form ax^{2} + bx + c = 0. Where a = 3 and c = 7. 
3x^{2}  6x  7 = 0 
Step 2: Move c, the constant term, to the righthand side of the equation. 
c = 7 3x^{2}  6x = 7 
Step 3: Factor out a from the lefthand side. This changes the value of the x coefficient. 
a = 3 3(x^{2}  2x) = 7 
Step 4: Complete the square of the expression in parentheses on the lefthand side of the equation.
The expression is x^{2}  2x. Divide the xcoefficient by two and square the result. 
x^{2}  2x x coefficient = 2 $\frac{2}{2}={}{1}\to {r}$ (1)^{2} = 1 
Step 5: Add the result from Step 4 to the parenthetical expression on the lefthand side. Then add a x result to the righthand side. To keep the equation true what is done to one side must also be done to the other. When adding the result to the parenthetical expression on the lefthand side the total value added is a x result. So this value must also be added to the righthand side. 
3(x^{2}  2x + 1) = 7 + 3(1) 
Step 6: 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 xcoefficient of the parenthetical expression divided by 2, as found in Step 4. 
3(x  1)^{2} = 10 
Now that the square has been completed, solve for x. 

Step 7: Divide both sides by a.

${\left(x1\right)}^{2}=\frac{10}{{3}}$ 
Step 8: 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. 
$x1=\pm \sqrt{\frac{10}{3}}$ 
Step 9: Solve for x. 
$x=1\pm \sqrt{\frac{10}{3}}$ 
Step 1: Write the equation in the general form ax^{2} + bx + c = 0. Where a = 5 and c = 0.6. 
5x^{2}  4x  0.6 = 0 
Step 2: Move c, the constant term, to the righthand side of the equation. 
c = 0.6 5x^{2}  4x = 0.6 
Step 3: Factor out a from the lefthand side. This changes the value of the xcoefficient. 
a = 5 5(x^{2}  0.8x) = 0.6 
Step 4: Complete the square of the expression in parentheses on the lefthand side of the equation.
The expression is x^{2}  0.8x. Divide the xcoefficient by two and square the result. 
x^{2}  0.8x xcoefficient = 0.8 $\frac{0.8}{2}={}{0.4}\to {r}$ (0.4)^{2} = 0.16 
Step 5: Add the result from Step 4 to the parenthetical expression on the lefthand side. Then add a x result to the righthand side. To keep the equation true what is done to one side must also be done to the other. When adding the result to the parenthetical expression on the lefthand side the total value added is a x result. So this value must also be added to the righthand side. 
5(x^{2}  0.8x + 0.16) = 0.6 + 5(0.16) 
Step 6: 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 xcoefficient of the parenthetical expression divided by 2 as found in Step 4. 
5(x  0.4)^{2} = 1.4 
Now that the square has been completed, solve for x. 

Step 7: Divide both sides by a.

${\left(x0.4\right)}^{2}=\frac{1.4}{{5}}=0.28$ 
Step 8: 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. 
$x0.4=\pm \sqrt{0.28}$ 
Step 9: Solve for x. 
$x=0.4\pm \sqrt{0.28}$ 
Related Links: Math algebra Factoring Quadratic Equations when a equals 1 Factoring Quadratic Equations when a ≠ 1 Algebra Topics 
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