It’s the case that factoring and using the quadratic formula are effective approaches to finding solutions of quadratic equations in Math.
There is also another method to solve such equations which is quadratic completing the square.
When we have a standard quadratic equation. ax^{\tt{2}} + bx + c = 0
Completing the square involves changing this equation into the following form.
a(x + k)^{\tt{2}} = h.
With a , b , c , h , k being real numbers, a ≠ 0.
From this altered form, we can then solve for x.
How to Solve by Completing the Square
In order get a quadratic equation in to vertex form, we follow a series of steps.
1) Make sure the x^{\tt{2}} co-efficient is 1.
2) Move the c term to the right hand side of the ‘=’ sign.
3) Observe the x co-effficient, half this number, and then square it.
4) Add the result to both sides of the equation, the left side should factor.
Then we have a 5th and 6th step in order to solve.
5) Take the square root of both sides.
6) Subtract from the left side leaving only x.
Ex
We can see these quadratic completing the square steps in action by considering
x^{\tt{2}} + 6x + 8 = 0.
The x^{\tt{2}} co-efficient is already 1.
– x^{\tt{2}} + 6x = {\text{-}}8
– x co-efficient is 6. [ \frac{6}{2} = 3, 32 = 9 ]
– x^{\tt{2}} + 6x + 9 = {\text{-}}8 + 9 => (x + 3)^{\tt{2}} = 1
– {\sqrt{(x + 3)^{\tt{2}}}} = \pm{\sqrt{1}}
We have a plus or minus sign as a square root can be a negative or positive number.
– x + 3 = \pm1
– x + 3 \space {\text{--}} \space 3 = \pm1 \space {\text{--}} \space 3 => x = {\text{-}}2 \space ,{\text{-}}4
Examples
(1.1)
Solve x^{\tt{2}} + 8x = 0 by completing the square.
Solution
Here the x^{\tt{2}} co-efficient is 1, but there is no c term. However that doesn’t matter, we can still proceed as usual.
x co-efficient is 8. \frac{8}{2} = 4, 42 = 16
x^{\tt{2}} + 8x + 16 = 16
(x + 4)^{\tt{2}} = 16
{\sqrt{(x + 4)^{\tt{2}}}} = \pm \space {\sqrt{16}}
x + 4 = \pm \space 4
x + 4 \space {\text{--}} \space 4 \space = \space \pm \space 4 \space {\text{--}} \space 4 => x = {\text{-}}8 \space , 0
(1.2)
Solve x^{\tt{2}} + 4x \space {\text{--}} \space 21 = 0 by completing the square.
Solution
There is a c term to initially move across.
x^{\tt{2}} + 4x = 21
x co-efficient is 4. \frac{4}{2} = 2, 22 = 4
x^{\tt{2}} + 4x + 4 = 21 + 4
(x + 2)^{\tt{2}} = 25
{\sqrt{(x + 2)^{\tt{2}}}} = \pm \space {\sqrt{25}}
x + 2 = \pm \space 5
x + 2 \space {\text{--}} \space 2 \space = \space \pm \space 5 \space {\text{--}} \space 2 => x = {\text{-}}7 \space , 3
Quadratic Completing the Square Solving,
Further Examples
Sometimes fractions can appear when dealing with completing the square examples.
Things can be a little bit more complicated than when dealing with only whole numbers, but the process of quadratic completing the square solving is the same.
(2.1)
Solve x^{\tt{2}} + 5x + 4 = 0 by completing the square.
Solution
The x^{\tt{2}} co-efficient is 1, we can move the c term over to the right, then proceed.
x^{\tt{2}} + 5x = {\text{-}}4
x co-efficient is 5. (\frac{5}{2})2 = \frac{25}{4}
x^{\tt{2}} + 5x + {\normalsize{\frac{25}{4}}} \space = \space {\text{-}}4 + {\normalsize{\frac{25}{4}}}
(x + {\normalsize{\frac{5}{2}}})^{\tt{2}} \space = \space {\normalsize{\frac{9}{4}}}
{\sqrt{(x + {\normalsize{\frac{5}{2}}})^{\tt{2}}}} \space = \space \pm \space {\sqrt{\frac{9}{4}}}
x + {\normalsize{\frac{5}{2}}} = \pm \space {\normalsize{\frac{3}{2}}}
x + {\normalsize{\frac{5}{2}}} \space {\text{--}} \space {\normalsize{\frac{5}{2}}} = \pm \space {\normalsize{\frac{3}{2}}} \space {\text{--}} \space {\normalsize{\frac{5}{2}}}
x = {\text{-}}{\normalsize{\frac{8}{2}}} \space , {\text{-}}{\normalsize{\frac{2}{2}}} => x = {\text{-}}4 \space , {\text{-}}1
(2.2)
Solve 3x^{\tt{2}} + 9x \space {\text{--}} \space 12 = 0 by completing the square.
Solution
Here initially the x^{\tt{2}} co-efficient is not 1.
We can make sure it is by dividing the equation through by the co-efficient, which is 3.
3x^{\tt{2}} + 9x \space {\text{--}} 12 = 0 \space \space \space ( \div 3)
x^{\tt{2}} + 3x \space {\text{--}} \space 4 = 0
x^{\tt{2}} + 3x = 4
x co-efficient is 3. (\frac{3}{2})2 = \frac{9}{4}
x^{\tt{2}} + 3x + {\normalsize{\frac{9}{4}}} \space = \space 4 + {\normalsize{\frac{9}{4}}}
(x + {\normalsize{\frac{3}{2}}})^{\tt{2}} \space = \space {\normalsize{\frac{25}{4}}}
{\sqrt{(x + {\normalsize{\frac{3}{2}}})^{\tt{2}}}} \space = \space \pm \space {\sqrt{\frac{25}{4}}}
x + {\normalsize{\frac{3}{2}}} = \pm \space {\normalsize{\frac{5}{2}}}
x + {\normalsize{\frac{3}{2}}} \space {\text{--}} \space {\normalsize{\frac{3}{2}}} = \pm \space {\normalsize{\frac{5}{2}}} \space {\text{--}} \space {\normalsize{\frac{3}{2}}}
x = {\text{-}}{\normalsize{\frac{8}{2}}} \space ,{\normalsize{\frac{2}{2}}} => x = {\text{-}}4 \space , 1
(2.3)
Solve {\text{-}}x^{\tt{2}} + 2x + 8 = 0 by completing the square.
Solution
Here initially the x^{\tt{2}} co-efficient is not 1.
We can make it so by multiplying the equation through by the co-efficient, which is –1.
{\text{-}}x^{\tt{2}} + 2x + 8 = 0 \space \space \space ( \times {\text{-}}1)
x^{\tt{2}} \space {\text{--}} \space 2x \space {\text{--}} \space 8 = 0
x^{\tt{2}} \space {\text{--}} \space 2x = 8
x co-efficient is –2. (\frac{{\text{-}}2}{2})2 = 1
x^{\tt{2}} \space {\text{--}} \space 2x + 1 \space = \space 8 + 1
(x \space {\text{--}} \space 1)^{\tt{2}} \space = \space 9
{\sqrt{(x \space {\text{--}} \space 1)^{\tt{2}}}} \space = \space \pm \space {\sqrt{9}}
x \space {\text{--}} \space 1 = \pm \space 3
x \space {\text{--}} \space 1 + 1 = \pm \space 3 + 1
x = {\text{-}}2 \space , 4