Completing the Square when a ≠ 1

A quadratic equation is an equation that contains a squared variable as its highest power on any variable. The general form of a quadratic equation is:

ax2 + bx + c = 0


Where a, b, and c are constants and a ≠ 0. In other words there must be a x2 term.

Some examples are:

x2 + 3x - 3 = 0
4x2 + 9 = 0 (Where b = 0)
x2 + 5x = 0 (where c = 0)

One way to solve a quadratic equation is by completing the square.

ax2 + bx + c = 0 → (x- r)2 = S


Where r and s are constants.

PART I of this topic focused on completing the square when a, the x2-coefficient, is 1. This part, PART II, will focus on completing the square when a, the x2-coefficient, is not 1.

Let's solve the following equation by completing the square:

2x2 + 8x - 5 = 0

Step 1: Write the equation in the general form

ax2 + bx + c = 0.


This equation is already in the proper form where a = 2 and c = -5.

2x2 + 8x - 5 = 0

Step 2: Move c, the constant term, to the right-hand side of the equation.

c = -5


2x2 + 8x = 5

Step 3: Factor out a from the left-hand side.


This changes the value of the x-coefficient.

a = 2


2(x2 + 4x) = 5

Step 4: Complete the square of the expression in parentheses on the left-hand side of the equation.


The expression is x2 + 4x.


Divide the x-coefficient by two and square the result.

x2 + 4x


x-coefficient = 4


4 2 =2r


(2)2 = 4

Step 5: Add the result from Step 4 to the parenthetical expression on the left-hand side. Then add a x result to the right-hand 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 left-hand side the total value added is a x result. So this value must also be added to the right-hand side.

2(x2 + 4x + 4) = 5 + 2(4)

Step 6: Rewrite the left-hand side as a perfect square and simplify the right-hand side.


When rewriting in perfect square format the value in the parentheses is the x-coefficient 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.

( x+2 ) 2 = 13 2

Step 8: Take the square root of both sides of the equation.


Remember that when taking the square root on the right-hand side the answer can be positive or negative.

x+2=± 13 2

Step 9: Solve for x.

x=2± 13 2

Example 1:      3x2 = 6x + 7

Step 1: Write the equation in the general form

ax2 + bx + c = 0.


Where a = 3 and c = -7.

3x2 - 6x - 7 = 0

Step 2: Move c, the constant term, to the right-hand side of the equation.

c = -7


3x2 - 6x = 7

Step 3: Factor out a from the left-hand side.


This changes the value of the x -coefficient.

a = 3


3(x2 - 2x) = 7

Step 4: Complete the square of the expression in parentheses on the left-hand side of the equation.


The expression is x2 - 2x.


Divide the x-coefficient by two and square the result.

x2 - 2x


x -coefficient = -2


2 2 =1r


(-1)2 = 1

Step 5: Add the result from Step 4 to the parenthetical expression on the left-hand side. Then add a x result to the right-hand 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 left-hand side the total value added is a x result. So this value must also be added to the right-hand side.

3(x2 - 2x + 1) = 7 + 3(1)

Step 6: Rewrite the left-hand side as a perfect square and simplify the right-hand side.


When rewriting in perfect square format the value in the parentheses is the x-coefficient 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.

( x1 ) 2 = 10 3

Step 8: Take the square root of both sides of the equation.


Remember that when taking the square root on the right-hand side the answer can be positive or negative.

x1=± 10 3

Step 9: Solve for x.

x=1± 10 3

Example 2:      5x2 - 0.6 = 4x

Step 1: Write the equation in the general form

ax2 + bx + c = 0.


Where a = 5 and c = 0.6.

5x2 - 4x - 0.6 = 0

Step 2: Move c, the constant term, to the right-hand side of the equation.

c = -0.6


5x2 - 4x = 0.6

Step 3: Factor out a from the left-hand side.


This changes the value of the x-coefficient.

a = 5


5(x2 - 0.8x) = 0.6

Step 4: Complete the square of the expression in parentheses on the left-hand side of the equation.


The expression is x2 - 0.8x.


Divide the x-coefficient by two and square the result.


x2 - 0.8x


x-coefficient = -0.8


0.8 2 =0.4r


(-0.4)2 = 0.16

Step 5: Add the result from Step 4 to the parenthetical expression on the left-hand side. Then add a x result to the right-hand 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 left-hand side the total value added is a x result. So this value must also be added to the right-hand side.

5(x2 - 0.8x + 0.16) = 0.6 + 5(0.16)

Step 6: Rewrite the left-hand side as a perfect square and simplify the right-hand side.


When rewriting in perfect square format the value in the parentheses is the x-coefficient 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.

( x0.4 ) 2 = 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 right-hand side the answer can be positive or negative.

x0.4=± 0.28

Step 9: Solve for x.

x=0.4± 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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