ORCCAOpen Resources for Community College Algebra

1. Use factor pairs.

   \begin{align*} x^2-2x-15\amp=0\\ (x-5)(x+3)\amp=0 \end{align*}
   \begin{align*} x-5\amp=0 \amp\text{ or }\amp\amp x+3\amp=0\\ x\amp=5 \amp\text{ or }\amp\amp x\amp=-3 \end{align*}

   So the solution set is \(\{5,-3\}\text{.}\)

2. Start by putting the equation in standard form and factoring out the greatest common factor.

   \begin{align*} 4x^2-40x\amp=-96\\ 4x^2-40x+96\amp=0\\ 4\left(x^2-10x+24\right)\amp=0\\ 4(x-6)(x-4)\amp=0 \end{align*}
   \begin{align*} x-6\amp=0 \amp\text{ or }\amp\amp x-4\amp=0\\ x\amp=6 \amp\text{ or }\amp\amp x\amp=4 \end{align*}

   So the solution set is \(\{4,6\}\text{.}\)

3. Use the AC method.

   \begin{align*} 6x^2+x-12\amp=0\\ \end{align*}

   Note that \(a\cdot c=-72\) and that \(\highlight{9\cdot-8}=-72\) and \(\highlight{9-8}=1\)

   \begin{align*} 6x^2\substitute{+9x-8x}-12\amp=0\\ \left(6x^2+9x\right)+\left(-8x-12\right)\amp=0\\ \highlight{3x}\left(2x+3\right)\mathbin{\highlight{-4}}\left(2x+3\right)\amp=0\\ \left(2x+3\right)\highlight{\left(3x-4\right)}\amp=0 \end{align*}
   \begin{align*} 2x+3\amp=0 \amp\text{ or }\amp\amp \highlight{3x-4}\amp=0\\ x\amp=-\frac{3}{2} \amp\text{ or }\amp\amp x\amp=\highlight{\frac{4}{3}} \end{align*}

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

4. Start by putting the equation in standard form.

   \begin{align*} (x-3)(x+2)\amp=14\\ x^2-x-6\amp=14\\ x^2-x-20\amp=0\\ (x-5)(x+4)\amp=0 \end{align*}
   \begin{align*} x-5\amp=0 \amp\text{ or }\amp\amp x+4\amp=0\\ x\amp=5 \amp\text{ or }\amp\amp x\amp=-4 \end{align*}

   So the solution set is \(\{5,-4\}\text{.}\)

5. Even though this equation has a power higher than \(2\text{,}\) we can still find all of its solutions by following the algorithm. Start by factoring out the greatest common factor.

   \begin{align*} x^3-64x\amp=0\\ x\left(x^2-64\right)\amp=0\\ x(x-8)(x+8)\amp=0 \end{align*}
   \begin{align*} x\amp=0 \amp\text{ or }\amp\amp x-8\amp=0 \amp\text{ or }\amp\amp x+8\amp=0\\ x\amp=0 \amp\text{ or }\amp\amp x\amp=8 \amp\text{ or }\amp\amp x\amp=-8 \end{align*}

   So the solution set is \(\{0,8,-8\}\text{.}\)

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

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

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 \(50\text{,}\) 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 \(32\text{,}\) 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.

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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 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 \(\left\{3+i\sqrt{3},3-i\sqrt{3}\right\}\text{.}\)