Solve by completing the square could take a little bit more time to do than solving by factoring. Keep reading to see a step-by-step explanation.
Before showing how to solve a quadratic equation by completing the square, you need to understand what a perfect square trinomial is and how to complete a square.
Perfect square trinomial = Binomial × same Binomial
Examples
(x + 4) × (x + 4) = x^{2} + 4x + 4x + 16 = x^{2} + 8x + 16 and x^{2} + 8x + 16 is a perfect square trinomial
(x + a) × (x + a) = x^{2} + ax + ax + a^{2} = x^{2} + 2ax + a^{2} and x^{2} + 2ax + a^{2} is a perfect square trinomial.
What is the relationship between the coefficient of the second term and the last term?
For x^{2} + 8x + 16, the coefficient of the second term is 8 and the last term is 16
(8/2)2 = 4^{2} = 16
For, x^{2} + 2ax + a^{2}, the coefficient of the second term is 2a and the last term is a^{2}
(2a/2)^{2} = a^{2}
So, what is the relationship?
The last term is obtained by dividing the coefficient of the second term by 2 and squaring the result.
In general, suppose x^{2} + bx is a binomial. You can complete the square by looking for a perfect square trinomial x^{2} + bx + c such that c = (b/2)^{2}
Now, let's say you have x^{2} + 20x and you want to find the last term to make the whole thing a perfect square trinomial.
Just do (20/2)^{2} = 10^{2}. The perfect square trinomial is x^{2} + 20x + 10^{2 }= (x + 10) × (x + 10)
The steps to follow are straightforward as you can see in the example shown below.
Example #1:
Solve by completing the square x^{2} + 6x + 8 = 0
x^{2} + 6x + 8 = 0
Subtract 8 from both sides of the equation so that you can isolate the two terms containing variables.
x^{2} + 6x + 8 - 8 = 0 - 8
x^{2} + 6x = - 8
To complete the square, always do the following 24 hours a day 365 days a year. It is never going to change when you solve by completing the square!
You are basically looking for a term to add to x^{2} + 6x that will make it a perfect square trinomial.
To this end, get the coefficient of the second term, divide it by 2 and raise it to the second power.
The second term is 6x and the coefficient is 6.
6/2 = 3 and after squaring 3, we get 3^{2}
Now, sometimes students forget to do this correctly. You need to add 3^{2} not only to the left side of the equation, but also to the right side of the equation.
x^{2} + 6x = - 8
Add 3^{2} to both sides of the equation above
x^{2} + 6x + 3^{2} = - 8 + 3^{2}
(x + 3)^{2} = -8 + 9
(x + 3)^{2} = 1
Take the square root of both sides so you end up with the linear equations.
√((x + 3)2) = √(1)
x + 3 = ±1
When x + 3 = 1, x = -2
When x + 3 = -1, x = -4
Example #2:
Solve by completing the square x^{2} + -6x + 8 = 0 instead of x^{2} + 6x + 8 = 0
The second term this time is -6x and the coefficient is -6.
-6/2 = -3 and after squaring -3, we get (-3)^{2} = 9
x^{2} + -6x = - 8
Add (-3)^{2} to both sides of the equation above
x^{2} + -6x + (-3)^{2} = - 8 + (-3)^{2}
(x + -3)^{2} = -8 + 9
(x + -3)^{2} = 1
Take the square root of both sides
√((x + -3)2) = √(1)
x + -3 = ±1
When x + -3 = 1, x = 4
When x + -3 = -1, x = 2
Example #3:
Solve by completing the square 3x^{2} + 8x + -3 = 0
3x^{2} + 8x + -3 = 0
Divide everything by 3. Always do that when the coefficient of the first term is not 1.
(3/3)x^{2}+ (8/3)x + -3/3 = 0/3
x^{2}+ (8/3)x + -1 = 0
Add 1 to both sides of the equation.
x^{2} + (8/3)x + -1 + 1 = 0 + 1
x^{2} + (8/3)x = 1
The second term is (8/3)x and the coefficient is 8/3.
8/3 ÷ 2 = 8/3 × 1/2 = 8/6 and after squaring 8/6, we get (8/6)^{2}
x^{2} + (8/3)x = 1
Add (8/6)^{2} to both sides of the equation above
x^{2} + (8/3)x + (8/6)^{2} = 1 + (8/6)^{2}
(x + 8/6)^{2} = 1 + 64/36
(x + 8/6)^{2} = 36/36 + 64/36 = (36 + 64)/36 = 100/36
Take the square root of both sides
√((x + 8/6)^{2}) = √(100/36)
x + 8/6 = ±10/6
x + 8/6 = 10/6
x = 10/6 - 8/6 = 2/6 = 1/3
x + 8/6 = - 10/6
x = -10/6 - 8/6 = -18/6 = -3
To solve by completing the square can become quickly hard as shown in example #3
Jun 06, 23 07:32 AM
May 01, 23 07:00 AM