Solving Quadratic Equations - Algebra II

To find the vertex of the parabola, you have to get it into vertex form:

The vertex can then be found at the coordinate .

To get to vertex form, we have to complete the square.

Move the 7 over to the other side by subtracting 7 from both sides of the equation:

You're going to have to add something to both sides of the equation...

...the question now is what. What number, when put in the box, would create a "perfect square" on the right-hand side of the equation?

Well, a perfect square trinomial is one whose factors are the same, like so:

In other words, we're looking for .

Well, if is what goes in the box, and is just , then must equal . Now we can solve for .

And since goes in the box, we need to add 4 to both sides:

Now we can factor the right-hand side very neatly:

After we clean up a bit...

...we get:

That gives us a vertex of .

To complete the square, the equation must be in the form:

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

Then, divide the middle coefficient by 2:

Square that and add it to both sides:

Now, you can factor the quadratic:

Take the square root of both sides:

Finish out the solution:

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

Then, divide the middle coefficient by 2:

Square that and add it to both sides:

Now, you can easily factor the quadratic:

Take the square root of both sides:

Finish out the solution:

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

Then, divide the middle coefficient by 2:

Square that and add it to both sides:

Now, you can easily factor the quadratic:

Take the square root of both sides:

Finish out the solution:

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

Then, divide the middle coefficient by 2:

Square that and add it to both sides:

Now, you can easily factor the quadratic:

Take the square root of both sides:

Finish out the solution:

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

Then, divide the middle coefficient by 2:

Square that and add it to both sides:

Now, you can easily factor the quadratic:

Take the square root of both sides:

Finish out the solution:

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

Then, divide the middle coefficient by 2:

Square that and add it to both sides:

Now, you can easily factor the quadratic:

Your next step would be to take the square root of both sides. At this point, however, you know that you cannot solve the problem. When you take the square root of both sides, you will be forced to take the square root of . This is impossible (at least in terms of real numbers), meaning that this problem must have no real solution.

From the original equation, we add 18 to both sides in order to set up our "completing the square."

To make completing the square sensible, we divide both sides by 2.

We now divide the x coefficient by 2, square the result, and add that to both sides.

Since the right side is now a perfect square, we can rewrite it as a square binomial.

Take the square root of both sides, simplify the radical and solve for x.

Vertex form for a parabola is

where (h, k) is the vertex.

We start by eliminating the leading coefficient by dividing both sides by 3.

We now subtract 6 from both sides to set up our "completing the square" technique.

To complete the square, we divide the x coefficient by 2, square the result, and add that result to both sides.

Since the right side is now a perfect square, we can rewrite it as a squared binomial.

Solve for y by adding 2 to both sides, then multiplying both sides by 3.

We start by moving the constant term of the quadratic to the other side of the equation, to set up the "completing the square" format.

Now to make completing the square sensible, we divide boths sides by 2 so that x^2 will not have a coefficient.

Now we can complete the square by dividing the x coefficient by 2 and squaring the result, then adding that result to both sides.

Because the left side is now a perfect square, we can rewrite it as a squared binomial.

Take the square root of both sides, and then solve for x.

Starting with the original equation, we move the constant term of the quadratic over to the other side, so we can set up our "completing the square."

Since the x^2 coefficient is already 1, we don't have to do any division. We can go straight to completing the square by dividing the x coefficient by 2, squaring the result, and adding that result to both sides.

Since the left side is now a perfect square, we can rewrite it as a squared binomial.

Now take the square root of both sides and solve for x.

We start by moving the constant term of the quadratic over to the other side of the equation, in order to set up our "completing the square" form.

Next we divide both sides by 4 so that the x^2 coefficient will be 1. That will allow us to complete the square.

Now we are ready to complete the square. We divide the x coefficient by 2, square the result, and add that result to both sides.

Because the right side is now a perfect square, we can rewrite it as a squared binomial.

To finish, all we have to do now is solve for y. We'll subtract 5 to both sides, then multiply by 4.

To complete the square, we need to have the x terms on one side and the numbers on the other. Therefore,

becomes

When we want to complete the square, we want an equation in the form or so that we can factor it into or . To do this, we take half of the numeric portion of what we want our b term to be (in this problem ) and square it, therefore:

Therefore, we add 16 to each side to obtain:

and

To complete the square, we need the left side in a form or so that we can factor it into form or .

To do this, we first divide out three on the left-hand side to obtain:

We then take 1/2 of the number in our term (in this case ) to obtain :

We then must add this to each side, but because we are completing the square inside of a parenthesis which is being multiplied by 3, we don't add 36 to each side, but rather 3 times 36, or 108. Therefore, we obtain:

and

Once the polynominal is factored out and everything is moved to the left, the equation becomes which does not factor evenly, so you could use the quadratic formula, or complete the square. To complete the square, a coefficient must be found to factor the polynominal into a perfect square. The polynominal factors to so we know that so and . To complete the square you add and subtract from the left side of the equation and strategically place parentheses to get , this simplifies to , which simplifies to ,

The first step is to make sure the x2 term has a coefficient of 1. Since we have that, we move onto the next step.

Next, move the "loose" number over to the other side.

Now divide the coefficient of the x term by 2 (don't forget the sign!). Add the square of this number to both sides.

Simplify:

Now the left side of the equation can be simplified to a squared factor. The factor that will be squared is going to be x plus the original x coefficient divided by 2 as we calculated above.

Take the square root of both sides:

Solve for x:

The first step when completing the square is to move the constant to the other side of the equation by subtracting from both sides. Don't forget the sign!

When completing the square, the lead coefficient should be one. To achieve this, divide both sides of the equation by the coefficient of the squared term.

and

This leaves you with the equation

a. Once the variables are on the left, the constant is on the right, and the lead coefficient is 1, you will create a perfect square trinomial on the left side of the equation. Do this by starting with the coefficient of the x term.

b. Divide this by 2.

c. Square this term.

d. Add the result to both sides of the equation.

e. The expression on the left side of the equation is now a perfect square trinomial and can be factored to: