Some quadratics are fairly simple to solve because they are of the form "something-with-x squared equals some number", and then you take the square root of both sides.
An example would be:
(x – 4)^{2} = 5
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Unfortunately, most quadratics don't come neatly squared like this. For your average everyday quadratic, you first have to use the technique of "completing the square" to rearrange the quadratic into the neat "(squared part) equals (a number)" format demonstrated above. For example:
First off, remember that finding the x-intercepts means setting y equal to zero and solving for the x-values, so this question is really asking you to "Solve 4x^{2} – 2x – 5 = 0".
Now, let's start the completing-the-square process. To begin, we have the original equation (or, if we had to solve first for "= 0", the "equals zero" form of the equation). In this case, we were asked for the x-intercepts of a quadratic function, which meant that we set the function equal to zero. So we're good to go. Our starting point is this equation:
4x^{2} – 2x – 5 = 0
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Now, contrary to everything we've learned before, we're going to move the constant (that is, the number that is not with a variable) over to the other side of the "equals" sign:
4x^{2} – 2x = 5
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When solving by completing the square, we'll want the x^{2} to be by itself, so we'll need to divide through by whatever is multiplied on this term. In this case, we've got a 4 multiplied on the x^{2}, so we'll need to divide through by 4 to get rid of this. Our result is:
Now we're going to do some work off on the side. Looking at the quadratic above, we have an x^{2} term and an x term on the left-hand side. We're going to work with the coefficient of the x term. In our present case, this value, along with its sign, is:
numerical coefficient: .
To created our completed square, we need to divide this numerical coefficient by 2 (or, which is the same thing, multiply it by one-half). In our case, we get:
derived value:
Now we'll square this derived value. (Of course, this will give us a positive number as a result.)
square of derived value:
Okay; now we go back to that last step before our diversion:
...and we add that "" to either side of the equation:
We can simplify the strictly-numerical stuff on the right-hand side:
At this point, we're ready to convert to completed-square form because, by adding that to either side, we had rearranged the left-hand side into a quadratic which is a perfect square. In other words, we can convert that left-hand side into a nice, neat squared binomial. But how?
The simplest way is to go back to the value we got after dividing by two (or, which is the same thing, multipliying by one-half), and using this, along with its sign, to form the squared binomial. In other words, in this case, we get:
Yay! Completed-square form! Now we can square-root either side (remembering the "plus-minus" on the strictly-numerical side):
Now we can solve for the values of the variable:
The "plus-minus" means that we have two solutions:
The solutions can also be written in rounded form as , or rounded to some reasonable number of decimal places (such as two).
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You will need probably rounded forms for "real life" answers to word problems, and for graphing. For instance, for the above exercise, it's a lot easier to graph an intercept at x = -0.9 than it is to try to graph the number in square-root form with a "minus" in the middle. But (warning!) in most other cases, you should assume that the answer should be in "exact" form, complete with all the square roots.
When you complete the square, make sure that you are careful with the sign on the numerical coefficient of the x-term when you multiply that coefficient by one-half. If you lose the sign from that term, you can get the wrong answer in the end because you'll forget which sign goes inside the parentheses in the completed-square form.
Also, don't be sloppy and wait to do the plus/minus sign until the very end. On your tests, you won't have the answers in the back to "remind" you that you "meant" to use the plus-minus, and you will likely forget to put the plus-minus into the answer. Besides, there's no reason to go ticking off your instructor by doing something wrong when it's so simple to do it right.
On the same note, make sure you draw in the square root sign, as necessary, when you square root both sides. Don't wait until the answer in the back of the book "reminds" you that you "meant" to put the square root symbol in there.
If you get in the habit of being sloppy, you'll only hurt yourself!
I'll do the same procedure as in the first exercise, in exactly the same order. (Study tip: Always working these problems in exactly the same way will help you remember the steps when you're taking your tests.)
First, I write down the equation they've given me.
x^{2} + 6x – 7 = 0
I move the constant term (the loose number) over to the other side of the "equals".
x^{2} + 6x = 7
The leading term is already only multiplied by 1, so I don't have to divide through by anything. So that step is done.
Now I'll grab some scratch paper, and do my computations. First, the coefficient of the "linear" term (that is, the term with just x, not the x^{2} term), with its sign, is:
numerical coefficient: +6
I'll multiply this by :
derived value:
My next step is to square this derived value:
square of derived value: (+3)^{2} = 9
Now I go back to my equation, and add this squared value to either side:
x^{2} + 6x + 9 = 7 + 9
I'll simplify the strictly-numerical stuff on the right-hand side:
x^{2} + 6x + 9 = 16
And now I'll convert the left-hand side to completed-square form, using the derived value (which I circled in my scratch-work, so I wouldn't lose track of it), along with its sign:
(x + 3)^{2} = 16
Now that the left-hand side is in completed-square form, I can square-root each side, remembering to put a "plus-minus" on the strictly-numerical side:
x + 3 = ± 4
...and then I'll solve for my two solutions:
x = –3 ±4
= –3 – 4, –3 + 4
= –7, 1
Then my answer is:
x = –7, 1
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Please take the time to work through the above two exercise for yourself, making sure that you're clear on each step, how the steps work together, and how I arrived at the listed answers. And then take the time to practice extra exercises from your book. With practice, this process can become fairly easy, especially if you're careful to work the exact same steps in the exact same order. Yes, "in real life" you'd use the Quadratic Formula or your calculator, but you should expect at least one question on the next test (and maybe the final) where you're required to show the steps for completing the square.
Note: Because the solutions to the second exercise above were integers, this tells you that we could have solved it by factoring.
x^{2} + 6x – 7 = 0
(x – 1)(x + 7) = 0
x – 1 = 0, x + 7 = 0
x = 1, x = – 7
Warning: If you are not consistent with remembering to put your plus/minus in as soon as you square-root both sides, then this is an example of the type of exercise where you'll get yourself in trouble. You'll write your answer for the second exercise above as "x = –3 + 4 = 1", and have no idea how they got "x = –7", because you won't have a square root symbol "reminding" you that you "meant" to put the plus/minus in. In other words, if you're sloppy, these easier problems will embarrass you!
On the next page, we'll do another example, and then show how the Quadratic Formula can be derived from the completing-the-square procedure...
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