Lane ORCCA (2020–2021): Open Resources for Community College Algebra

To solve the quadratic equation \(x^2+6x=16\text{,}\) on the left side we can complete the square by adding \(\left(\frac{b}{2}\right)^2\text{;}\) note that \(b=6\) in this case, which makes \(\left(\frac{b}{2}\right)^2=\left(\frac{6}{2}\right)^2=3^2=9\text{.}\) We add it to both sides to maintain equality.

Now that we have completed the square, we can solve the equation using the square root property.

The solution set is \(\{-8, 2\}\text{.}\)

We will solve \(x^2-14x+11=0\text{.}\) We see that the polynomial on the left side is not a perfect square trinomial, so we need to complete the square. We subtract \(11\) from both sides so we can add the missing term on the left.

Next comes the completing-the-square step. We need to add the correct number to both sides of the equation to make the left side a perfect square. Remember that Fact 11.2.1 states that we need to use \(\left(\frac{b}{2}\right)^2\) for this. In our case, \(b=-14\text{,}\) so \(\left(\frac{b}{2}\right)^2=\left(\frac{-14}{2}\right)^2=49\)

The solution set is \(\{7-\sqrt{38}, 7+\sqrt{38}\}\text{.}\)

To complete the square, we will first move the constant term to the right side of the equation. Then we will use Fact 11.2.1 to find \(\left(\frac{b}{2}\right)^2\) to add to both sides.

In our case, \(b=-20\text{,}\) so \(\left(\frac{b}{2}\right)^2=\left(\frac{-20}{2}\right)^2=100\)

The solution set is \(\{-1, 21\}\text{.}\)

We will first move the constant term to the right side of the equation:

Next, to complete the square, we will need to find the right number to add to both sides. According to Fact 11.2.1, we need to divide the value of \(b\) by \(2\) and then square the result to find the right number. First, divide by \(2\text{:}\)

and then we square that result:

Now we can add the \(\frac{9}{4}\) from Equation (11.2.2) to both sides of the equation to complete the square.

Now, to factor the seemingly complicated expression on the left, just know that it should always factor using the number from the first step in the completing the square process, Equation (11.2.1).

The solution set is \(\{-2,5\}\text{.}\)

Because there is a leading coefficient of \(2\text{,}\) we will divide both sides by \(2\text{.}\)

Next, we will complete the square. Since \(b=1\text{,}\) first,

and next, squaring that, we have

So we will add \(\frac{1}{4}\) from Equation (11.2.4) to both sides of the equation:

Here, remember that we always factor with the number found in the first step of completing the square, Equation (11.2.3).

The solution set is \(\left\{\frac{-1-\sqrt{7}}{2}, \frac{-1+\sqrt{7}}{2}\right\}\)

The formula is in the form \(x^2+bx\text{,}\) so we will need to add \(\left(\frac{b}{2}\right)^2\) to complete the square by Fact 11.2.1. When we had an equation, we could add the same quantity to both sides. But now we do not wish to change the left side, since we are trying to end up with a formula that still says \(q(x)=\ldots\text{.}\) Instead, we will add and subtract the term from the right side in order to maintain equality. In this case,

To maintain equality, we will both add and subtract \(16\) on the same side of the equation. It is functionally the same as adding \(0\) on the right, but the \(16\) makes it possible to factor the expression in a particular way:

Now that we have completed the square, our function is in vertex form. The vertex is \((-4,-16)\text{.}\) One way to verify that our work is correct is to graph the original version of the function and check that the vertex is where it should be.

To complete the square, we need to add and subtract \(\left(-\frac{12}{2}\right)^2=(-6)^2=36\) on the right side.

The vertex is \((6,-33)\text{.}\)

Before we can complete the square, we will factor the \(\highlight{5}\) out of the first two terms.

Now we will complete the square inside the parentheses by adding and subtracting \(\left(\frac{4}{2}\right)^2=2^2=4\text{.}\)

Notice that the constant that we subtracted is inside the parentheses, but it will not be part of our perfect square trinomial. In order to bring it outside, we need to multiply it by \(5\text{.}\) We are distributing the \(5\) to that term so we can combine it with the outside term.

The vertex is \((-2,5)\text{.}\)

First, we will factor the leading coefficient out of the first two terms.

Next, we will complete the square for \(x^2+\frac{4}{3}x\) inside the grouping symbols by adding and subtracting the right number. To find that number, we divide the value of \(b\) by two and square the result. That looks like:

and then,

Adding and subtracting the value from Equation (11.2.8), we have:

The vertex is \(\left(-\frac{2}{3},-\frac{5}{12}\right)\text{.}\)

To find the maximum is essentially the same as finding the vertex, which we can find by completing the square. To complete the square for \(I(x)=-100x^2+1000x+20000\text{,}\) we start by factoring out the \(-100\) from the first two terms:

The vertex is the point \((5,22500)\text{.}\) This implies Tyrone should raise the price per painting \(\substitute{5}\) times, which is \(\substitute{5}\cdot20=100\) dollars. He would sell \(100-5(\substitute{5})=75\) paintings. This would make the price per painting \(200+100=300\) dollars, and his annual income from paintings would become \(\$22500\) by this model.

To start, we'll note that this function opens upward because the leading coefficient, \(2\text{,}\) is positive.

Now we will complete the square to find the vertex. We will factor the \(2\) out of the first two terms, and then add and subtract \(\left(\frac{2}{2}\right)^2=1^2=\highlight{1}\) on the right side.

The vertex is \((-1,-8)\) so the axis of symmetry is the line \(x=-1\text{.}\)

To find the \(y\)-intercept, we'll replace \(x\) with \(0\) or read the value of \(c\) from the function in standard form:

The \(y\)-intercept is \((0,-6)\) and we will find its symmetric point on the graph, which is \((-2,-6)\text{.}\)

Next, we'll find the horizontal intercepts. We see this function factors so we will write the factored form to get the horizontal intercepts.

The \(x\)-intercepts are \((1,0)\) and \((-3,0)\text{.}\)

Now we will plot all of the key points and draw the parabola.

In this function, the leading coefficient is negative so it will open downward. To complete the square we first factor \(-1\) out of the first two terms.

Now, we add and subtract the correct number on the right side of the function: \(\left(\frac{b}{2}\right)^2=\left(\frac{4}{2}\right)^2=2^2=\highlight{4}\text{.}\)

The vertex is \((-2,3)\) so the axis of symmetry is the line \(x=-2\text{.}\)

We find the \(y\)-intercept by looking at the value of \(c\text{,}\) which is \(-1\text{.}\) So, the \(y\)-intercept is \((0,-1)\) and we will find its symmetric point on the graph, \((-4,-1)\text{.}\)

The original expression, \(-x^2-4x-1\text{,}\) does not factor so to find the \(x\)-intercepts we need to set \(p(x)=0\) and complete the square or use the quadratic formula. Since we just went through the process of completing the square above, we can use that result to save several repetitive steps.

The \(x\)-intercepts are approximately \((-3.7,0)\) and \((-0.3,0)\text{.}\) Now we can plot all of the points and draw the parabola.