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

To estimate \(\sqrt{28}\text{,}\) we can find the nearest perfect squares that are whole numbers on either side of \(28\text{.}\) Recall that the perfect squares are \(1, 4, 9, 16, 25, 36, 49, 64,\dots\) The perfect square that is just below \(28\) is \(25\) and the perfect square just above \(28\) is \(36\text{.}\) This tells us that \(\sqrt{28}\) is between \(\sqrt{25}\) and \(\sqrt{36}\text{,}\) or between \(5\) and \(6\text{.}\) We can also say that \(\sqrt{28}\) is closer to \(5\) than \(6\) because \(28\) is closer to \(25\text{,}\) so we think \(5.2\) or \(5.3\) would be a good estimate.

On the calculator we can see that \(\sqrt{28}\approx5.29\text{,}\) so our guess was very close to reality.

1. \(\displaystyle \begin{aligned}[t] \sqrt{18}\cdot\sqrt{2}\amp=\sqrt{18\cdot2}\\ \amp=\sqrt{36}\\ \amp=6\end{aligned} \)

2. \(\displaystyle \begin{aligned}[t] \frac{\sqrt{18}}{\sqrt{2}}\amp=\sqrt{\frac{18}{2}}\\ \amp=\sqrt{9}\\ \amp=3\end{aligned} \)

Recall that the perfect squares are \(1, 4, 9, 16, 25, 36, 49, 64,\dots\) To simplify the \(\sqrt{54}\text{,}\) we need to look at that list and find the largest perfect square the goes into \(54\) evenly. In this case, it is \(\highlight{9}\text{.}\) We then break up \(54\) into two factors \(\highlight{9}\) and \(6\text{,}\) and we have:

Since \(6\) has no perfect square factors, we can stop.

Note that \(25\) is a perfect-square factor of \(50\) and that \(9\) is a perfect-square factor of \(27\text{.}\) Now we have:

Recall that radicals can only be added if the radicands match identically, so we cannot initially combine these two terms. However, if we simplify first, we may be able to add terms later. Note that \(16\) is a perfect-square factor of \(32\) and that \(25\) is a perfect-square factor of \(50\text{.}\)

We will use the FOIL method to expand this expression:

It's important here to suppress any urge you may have to expand the squared binomial. We begin by isolating the squared expression.

Now that we have the squared expression isolated, we can use the square root property.

The solution set is \(\left\{2\sqrt{2}+2,-2\sqrt{2}+2\right\}\text{.}\)

Let's start by representing the length of the triangle, measured in inches, by the letter \(x\text{.}\) That would also make the other side \(x\) inches long.

Faven should now set up the Pythagorean theorem regarding the picture. That would be

Solving this equation, we have:

Faven should make the sides of her triangle about \(11.3\) inches long to force the hypotenuse to be \(16\) inches long.

1. First we should change the equation into standard form.

   \begin{align*} x^2+4x\amp=6\\ x^2+4x-6\amp=0 \end{align*}

   Next, we check and see that we cannot factor the left side or use the square root property so we must use the quadratic formula. We identify that \(\substitute{a=1}\text{,}\) \(\substitute{b=4}\text{,}\) and \(\substitute{c=-6}\text{.}\) We will substitute them into the quadratic formula:

   \begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-\substitute{4}\pm\sqrt{(\substitute{4})^2-4(\substitute{1})(\substitute{-6})}}{2(\substitute{1})}\\ \amp=\frac{-4\pm\sqrt{16+24}}{2}\\ \amp=\frac{-4\pm\sqrt{40}}{2}\\ \amp=\frac{-4\pm\sqrt{\highlight{4}\cdot10}}{2}\\ \amp=\frac{-4\pm\sqrt{\highlight{4}}\cdot\sqrt{10}}{2}\\ \amp=\frac{-4\pm\highlight{2}\sqrt{10}}{2}\\ \amp=-\frac{4}{2}\pm\frac{2\sqrt{10}}{2}\\ \amp=-2\pm\sqrt{10} \end{align*}

   So the solution set is \(\left\{-2+\sqrt{10},-2-\sqrt{10}\right\}\text{.}\)

2. Since the equation \(5x^2-2x+1=0\) is already in standard form, we check and see that we cannot factor the left side or use the square root property so we must use the quadratic formula. We identify that \(\substitute{a=5}\text{,}\) \(\substitute{b=-2}\text{,}\) and \(\substitute{c=1}\text{.}\) We will substitute them into the quadratic formula:

   \begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-\substitute{(-2)}\pm\sqrt{\substitute{(-2)}^2-4(\substitute{5})(\substitute{1})}}{2(\substitute{5})}\\ \amp=\frac{2\pm\sqrt{4-20}}{10}\\ \amp=\frac{2\pm\sqrt{-16}}{10} \end{align*}

   Since the solutions have square roots of negative numbers, we must conclude that there are no real solutions.

1. \(2(x-3)^2-5x=6\) is quadratic.

2. \(2(x-3)-5x=6\) is linear.

3. \(2x-6=7x^3\) is neither linear or quadratic.

4. \(2x^2-6=7x^2\) is quadratic.

5. \(2\sqrt{x}-x-6=0\) is neither linear or quadratic.

6. \(2x-(x-6)=0\) is linear.

1. To solve the equation \(4x^2-11x+6=0\) we first note that it is quadratic. Since there is a linear term (\(-11x\)), we must use either factoring or the quadratic formula, and we will try factoring first. Since the leading coefficient is \(4\text{,}\) we will try the AC method. In this case, \(ac=24\text{:}\) numbers that multiply to be \(24\) and add to be \(-11\) are \(-8\) and \(-3\text{.}\) So we split up th equation like this:

   \begin{align*} 4x^2-11x+6\amp=0\\ 4x^2\mathbin{\highlight{-}}\mathbin{\highlight{8x-3x}}+6\amp=0\\ \left(4x^2\mathbin{\highlight{-}}\mathbin{\highlight{8x}}\right)+(\mathbin{\highlight{-}}\mathbin{\highlight{3x}}+6)\amp=0\\ 4x(x-2)-3(x-2)\amp=0\\ (x-2)(4x-3)\amp=0 \end{align*}
   \begin{align*} x-2\amp=0 \amp \text{ or } \amp\amp 4x-3\amp=0\\ x\amp=2 \amp \text{ or } \amp\amp x\amp=\frac{3}{4} \end{align*}

   So, the solution set is \(\left\{2,\frac{3}{4}\right\}\text{.}\)

2. To solve the equation \(2(x-6)^2-16=0\) we first note that it is quadratic. Since there is no linear term, we should try using the square root method.

   \begin{align*} 2(x-6)^2-16\amp=0\\ 2(x-6)^2\amp=16\\ (x-6)^2\amp=8 \end{align*}
   \begin{align*} x-6\amp=\sqrt{8} \amp \text{ or } \amp\amp x-6\amp=-\sqrt{8}\\ x-6\amp=\sqrt{\highlight{4}\cdot2} \amp \text{ or } \amp\amp x-6\amp=-\sqrt{\highlight{4}\cdot2}\\ x-6\amp=\sqrt{\highlight{4}}\cdot\sqrt{2} \amp \text{ or } \amp\amp x-6\amp=-\sqrt{\highlight{4}}\cdot\sqrt{2}\\ x-6\amp=\highlight{2}\sqrt{2} \amp \text{ or } \amp\amp x-6\amp=-\highlight{2}\sqrt{2}\\ x\amp=6+\highlight{2}\sqrt{2} \amp \text{ or } \amp\amp x\amp=6-\highlight{2}\sqrt{2} \end{align*}

   So, the solution set is \(\left\{6+2\sqrt{2},6-2\sqrt{2}\right\}\text{.}\)

3. To solve the equation \(2(x-6)-16=0\) we first note that it is linear. Since it is linear, we just need to isolate the terms with the variable on one side and all the other terms on the other side of the equals sign.

   \begin{gather*} 2(x-6)-16=0\\ 2x-12-16=0\\ 2x-28=0\\ 2x=28\\ x=14 \end{gather*}

   So, the solution set is \(\{14\}\text{.}\)

4. To solve the equation \(2(x-6)^2-15x=0\) we first note that it is quadratic. Since there is a linear term, we must use either factoring or the quadratic formula. Before we can decide which to use, we need to put the equation in standard form:

   \begin{align*} 3(x-4)^2-15x=0\\ 3(x-4)(x-4)-15x=0\\ 3(x^2-8x+16)-15x=0\\ 3x^2-24x+48-15x=0\\ 3x^2-39x+48=0\\ \end{align*}

   Now we can see that the left hand side does not factor easily, so we will fall back on the quadratic formula. We identify that \(\substitute{a=3}\text{,}\) \(\substitute{b=-39}\text{,}\) and \(\substitute{c=48}\text{.}\)

   \begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-\substitute{(-39)}\pm\sqrt{\substitute{(-39)}^2-4(\substitute{3})(\substitute{48})}}{2(\substitute{3})}\\ \amp=\frac{39\pm\sqrt{1521-576}}{6}\\ \amp=\frac{39\pm\sqrt{945}}{6}\\ \amp=\frac{39\pm\sqrt{\highlight{9}\cdot105}}{6}\\ \amp=\frac{39\pm\sqrt{\highlight{9}}\cdot\sqrt{105}}{6}\\ \amp=\frac{39\pm3\sqrt{105}}{6} \end{align*}

   So the solution set is \(\left\{\frac{39+3\sqrt{105}}{6},\frac{39-3\sqrt{105}}{6}\right\}\text{.}\)

Start by splitting the \(-1\) from the \(12\) and by looking for the largest perfect-square factor of \(-12\text{,}\) which happens to be \(4\text{.}\)

There is no \(m\) term so we will use the square root method.

The solution set is \(\left\{-\highlight{2}\lighthigh{i}\sqrt{2},\highlight{2}\lighthigh{i}\sqrt{2}\right\}\text{.}\)

So, the solution set is \(\left\{2+\highlight{3}\lighthigh{i}\sqrt{2},2-\highlight{3}\lighthigh{i}\sqrt{2}\right\}\text{.}\)

All three of the equations here are very similar, so we will need to examine them closely to choose the best method for solving them.

1. To solve the equation \((x-4)^2-2=0\text{,}\) first note that there is no linear term: there is only a square and a constant. This leads us to consider the square root method. Before doing that, isolate the square:

   \begin{align*} (x-4)^2-2\amp=0\\ (x-4)^2\amp=2 \end{align*}

   Now we can apply the square root method to the equation.

   \begin{align*} x-4\amp=\sqrt{2}\amp\text{ or }\amp\amp x-4\amp=-\sqrt{2}\\ x\amp=4+\sqrt{2}\amp\text{ or }\amp\amp x\amp=4-\sqrt{2} \end{align*}

   So the solution set is \(\left\{4+\sqrt{2},4-\sqrt{2}\right\}\)

2. To solve the equation \((x-4)^2-2x=0\text{,}\) first note that there is a linear term (\(-2x\)), so we must use either factoring or the quadratic formula. To use either, we must first put the equation in standard form.

   \begin{align*} (x-4)^2-2x\amp=0\\ (x-4)(x-4)-2x\amp=0\\ x^2-8x+16-2x\amp=0\\ x^2-10x+16\amp=0 \end{align*}

   Now that the equation is in standard form, we can decide whether to use factoring or the quadratic formula. While the quadratic formula always works, it can take more time than factoring if factoring is possible. In this case, factoring entails answering the question “are there two integers that multiply to be \(16\) and add to be \(-10\text{?}\)” The answer is “yes” : \(-8\) and \(-2\) are such numbers.

   \begin{align*} x^2-10x+16\amp=0\\ (x-8)(x-2)\amp=0 \end{align*}
   \begin{align*} x-8\amp=0\amp\text{ or }\amp\amp x-2\amp=0\\ x\amp=8\amp\text{ or }\amp\amp x\amp=2 \end{align*}

   So the solution set is \(\{2,8\}\text{.}\)

3. To solve the equation \((x-4)^2+2x=0\text{,}\) first note that there is a linear term (\(+2x\)), so we must use either factoring or the quadratic formula. To use either, we must first put the equation in standard form.

   \begin{align*} (x-4)^2+2x\amp=0\\ (x-4)(x-4)+2x\amp=0\\ x^2-8x+16+2x\amp=0\\ x^2-6x+16\amp=0 \end{align*}

   Now that the equation is in standard form, we can decide whether to use factoring or the quadratic formula. In this case, factoring entails answering the question “are there two integers that multiply to be \(16\) and add to be \(-6\text{?}\)” The answer is “no,” so we must use the quadratic formula. First, identify that \(\substitute{a=1}\text{,}\) \(\substitute{b=-6}\text{,}\) and \(\substitute{c=16}\text{.}\)

   \begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(\substitute{-6})\pm\sqrt{(\substitute{-6})^2-4(\substitute{1})(\substitute{16})}}{2(\substitute{1})}\\ \amp=\frac{6\pm\sqrt{36-48}}{2}\\ \amp=\frac{6\pm\sqrt{-12}}{2}\\ \end{align*}

   At this point, we notice that the solutions are complex. Continue to simplify until they are completely reduced.

   \begin{align*} x\amp=\frac{6\pm\sqrt{\highlight{4}\cdot\lighthigh{-1}\cdot3}}{2}\\ \amp=\frac{6\pm\sqrt{\highlight{4}}\cdot\sqrt{\lighthigh{-1}}\cdot\sqrt{3}}{2}\\ \amp=\frac{6\pm\highlight{2}\lighthigh{i}\sqrt{3}}{2}\\ \amp=\frac{6}{2}\pm\frac{2i\sqrt{3}}{2}\\ \amp=3\pm i\sqrt{3} \end{align*}

   So the solution set is \(\left\{3+i\sqrt{3},3-i\sqrt{3}\right\}\text{.}\)

To find the vertex of a parabola algebraically, we use the formula \(h=-\frac{b}{2a}\) to find the axis of symmetry first. For our equation, \(\substitute{a=3}\) and \(\lighthigh{b=8}\text{,}\) so:

Next, we evaluate \(f\mathopen{}\left(\substitute{-\frac{4}{3}}\right)\mathclose{}\) to find the \(y\)-coordinate of the vertex.

So, the parabola has its axis of symmetry at \(x=-\frac{4}{3}\) and has its vertex at the point \(\left(-\frac{4}{3},-\frac{37}{3}\right)\text{.}\)

To determine the vertex of \(g(x)=2x^2+8x+2\text{,}\) we want to find the \(x\)-value of the vertex first. We will use \(h=-\frac{b}{2a}\) for \(\substitute{a=2}\) and \(\lighthigh{b=8}\text{:}\)

To find the vertex, we evaluate \(g(\substitute{-2})\text{.}\)

Now we know that our axis of symmetry is the line \(x=-2\) and the vertex is the point \((-2,-6)\text{.}\) We will set up our table with two values on each side of \(x=-2\text{.}\) We choose \(x= -4, -3, \highlight{-2}, -1\text{,}\) and \(0\text{.}\) Then, we'll determine the \(y\)-coordinates by replacing \(x\) with each value and we have the complete table as shown in Table 9.8.23. Notice that each pair of \(y\)-values on either side of the vertex match. This helps us to check that our vertex and \(y\)-values are correct.

Now that we have our table, we will plot the points and draw in the axis of symmetry as shown in Figure 9.8.24. We complete the graph by drawing a smooth curve through the points and drawing an arrow on each end as shown in Figure 9.8.25

1. To find the vertical intercept, we evaluate the function when \(x=0\text{.}\) \(h(\substitute{0})=6(\substitute{0})^2-13(\substitute{0})+6=6\text{.}\) So, the vertical intercept of \(h\) is \((0,6)\text{.}\)

   Next, to find the horizontal intercepts, we set the function equal to zero. To solve that equation we will use Process 9.5.1. For this particular example, we will practice factoring using the AC method. Here, \(ac=36\) and factor pairs that add up to \(-13\) are \(-9\) and \(-4\text{,}\) as you will see.

   \begin{align*} \substitute{h(x)}\amp=6x^2-13x+6\\ \substitute{0}\amp=6x^2\mathbin{\highlight{-}}\mathbin{\highlight{13x}}+6\\ 0\amp=6x^2\mathbin{\highlight{-}}\mathbin{\highlight{9x}}\mathbin{\highlight{-}}\mathbin{\highlight{4x}}+6\\ 0\amp=3x\highlight{(2x-3)}-2\highlight{(2x-3)}\\ 0\amp=(3x-2)\highlight{(2x-3)} \end{align*}
   \begin{align*} 0\amp=3x-2\amp\text{ or }\amp\amp0\amp=2x-3\\ x\amp=\frac{2}{3}\amp\text{ or }\amp\amp x\amp=\frac{3}{2} \end{align*}

   So, the horizontal intercepts are \(\left(\frac{2}{3},0\right)\) and \(\left(\frac{3}{2},0\right)\text{.}\)

2. To find the vertical intercept, we have to evaluate the function when \(x=0\text{.}\) \(k(\substitute{0})=2(\substitute{0})^2-2(\substitute{0})-5=-5\text{.}\) So, the vertical intercept of \(k\) is \((0,-5)\text{.}\)

   Next, to find the horizontal intercepts, we set the function equal to zero. To solve that equation we will use Process 9.5.1. For this particular example, we will use the quadratic formula because the square root method will not work (because there is a linear term) and factoring fails.

   \begin{align*} \substitute{k(x)}\amp=2x^2-2x-5\\ \substitute{0}\amp=2x^2-2x-5\\ \end{align*}

   We identify that \(\substitute{a=2}\text{,}\) \(\lighthigh{b=-2}\text{,}\) and \(c=-5\)

   \begin{align*} x\amp=\frac{-\lighthigh{b}\pm\sqrt{\lighthigh{b}^2-4\substitute{a}c}}{2\substitute{a}}\\ x\amp=\frac{-(\lighthigh{-2})\pm\sqrt{(\lighthigh{-2})^2-4(\substitute{2})(-5)}}{2(\substitute{2})}\\ x\amp=\frac{2\pm\sqrt{4+40}}{4}\\ x\amp=\frac{2\pm\sqrt{44}}{4}\\ x\amp=\frac{2\pm\highlight{2}\sqrt{11}}{4}\\ x\amp=\frac{2}{4}\pm\frac{2\sqrt{11}}{4}\\ x\amp=\frac{1}{2}\pm\frac{\sqrt{11}}{2}\\ x\amp=\frac{1\pm\sqrt{11}}{2} \end{align*}

   So, the horizontal intercepts are \(\left(\frac{1+\sqrt{11}}{2},0\right)\) and \(\left(\frac{1-\sqrt{11}}{2},0\right)\text{.}\) If you wanted to graph these points, you would need to approximate them as \(\left(2.16,0\right)\) and \(\left(-1.16,0\right)\text{.}\)

To start, we'll note that this function will open downward, as the leading coefficient is negative.

To find the \(y\)-intercept, we'll evaluate \(j(0)\text{:}\) \(j(\substitute{0})=-2(\substitute{0})^2+6(\substitute{0})+8=8\text{.}\) The \(y\)-intercept is \((0,8)\text{.}\)

Next, we'll find the horizontal intercepts by setting \(j(x)=0\) and solving for \(x\text{:}\)

The \(x\)-intercepts are \((4,0)\) and \((-1,0)\text{.}\)

Lastly, we'll determine the vertex. Noting that \(\substitute{a=-2}\) and \(\lighthigh{b=6}\text{,}\) we have:

Using the \(x\)-value \(\substitute{1.5}\) to find the \(y\)-coordinate, we have: \(j(\substitute{1.5})=-2(\substitute{1.5})^2+6(\substitute{1.5})+8=12.5\text{.}\) The vertex is the point \((1.5,12.5)\text{,}\) and the axis of symmetry is the line \(x=1.5\text{.}\)

We're now ready to graph this function. We'll start by drawing and scaling the axes so all of our key features will be displayed as shown in Figure 9.8.28. Next, we'll plot these key points as shown in Figure 9.8.29. Finally, we'll note that this parabola opens downward and connect these points with a smooth curve, as shown in Figure 9.8.30.

1. The vertical intercept is found when \(\substitute{x=0}\text{.}\) \(h(\substitute{0})=0.0018(\substitute{0})^2-0.9(\substitute{0})=0\text{.}\) So the vertical intercept is \((0,0)\) which means that the ball was released \(0\) meters from the rim (on the rim) at a height of \(0\) meters above the rim (again, on the rim).

2. The horizontal intercepts are found when \(h(x)=0\text{.}\) This will be a quadratic equation that we can solve using factoring.

   \begin{align*} h(x)\amp=0\\ 0.0018x^2-0.9x\amp=0\\ x\left(0.0018x-0.9\right)\amp=0 \end{align*}
   \begin{align*} x\amp=0\amp\text{ or }\amp\amp 0.0018x-0.9\amp=0\\ x\amp=0\amp\text{ or }\amp\amp 0.0018x\amp=0.9\\ x\amp=0\amp\text{ or }\amp\amp x\amp=500 \end{align*}

   The horizontal intercepts are \((0,0)\) and \((500,0)\text{.}\) The interpretation of \((0,0)\) is the same as before. The interpretation of \((500,0)\) is that the other side of the rim of the dish is \(500\) meters away. If the ball were to somehow roll all the way down, and then back up the other side, it would pop up at the opposite rim exactly \(500\) meters horizontally away.

3. The vertex is found when the \(x\)-value is at the axis of symmetry. To find that, we use the formula \(x=-\frac{\lighthigh{b}}{2\substitute{a}}\) where \(\substitute{a=0.0018}\) and \(\lighthigh{b=-0.9}\)

   \begin{align*} x\amp=-\frac{\lighthigh{b}}{2\substitute{a}}\\ x\amp=-\frac{\lighthigh{-0.9}}{2(\substitute{0.0018})}\\ \amp=250 \end{align*}

   Then, to find the \(y\)-value, we substitute that \(x\)-value, \(\substitute{250}\) into the original function.

   \begin{align*} h(\substitute{250})\amp=0.0018(\substitute{250})^2-0.9(\substitute{250})\\ \amp=-112.5 \end{align*}

   The vertex is the point \((250,-112.5)\text{.}\) Since the vale of \(a\) is positive, we know that this parabola opens upward. This means that the vertex is the lowest point on the graph. So the lowest point of the telescope dish is \(250\) meters from the rim (horizontally) and \(112.5\) meters below the height of the rim.

Recall that Definition 9.6.1 says that a quadratic function has the form \(f(x)=ax^2+bx+c\) where \(a\text{,}\) \(b\text{,}\) and \(c\) are real numbers, and \(a \neq 0\text{.}\)

1. The equation \(y-1=3(x-2)-5\) is not quadratic. It is in fact a linear equation.

2. The equation \(y=3(x-2)^2-5\) is quadratic. We could simplify the right hand side and would have something of the form \(y=ax^2+bx+c\text{.}\)

3. The equation \(y=\sqrt{x-2}-5\) is not quadratic. The \(x\) is inside a radical, not squared, so it cannot be converted into the form \(y=ax^2+bx+c\text{.}\)

4. The equation \(y^2=x^2-5\) is not quadratic. It cannot be re-written in the form \(y=ax^2+bx+c\) (due to the \(y^2\) term).