Open Resources for Community College Algebra

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.

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 \(\left\{-\firsthighlight{2}\secondhighlight{i}\sqrt{2},\firsthighlight{2}\secondhighlight{i}\sqrt{2}\right\}\text{.}\)

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

1. Since the variable \(x\) only appears once, we can apply steps one at a time to undo all of the operations that are done to \(x\) and eventually isolate it.

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

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

2. Since the variable \(x\) only appears once, we can apply steps one at a time to undo all of the operations that are done to \(x\) and eventually isolate it.

   \begin{align*} \sqrt{3x+2}-2\amp=5\\ \sqrt{3x+2}\amp=7\\ \left(\sqrt{3x+2}\right)^2\amp=7^2\\ 3x+2\amp=49\\ 3x\amp=47\\ x\amp=\frac{47}{3} \end{align*}

   At this point \(\frac{47}{3}\) is only a potential solution. We may have introduced an extraneous solution at the point where we squared both sides. So we should check it.

   \begin{align*} \sqrt{3\cdot\substitute{\frac{47}{3}}+2}-2\amp\stackrel{?}{=}5\\ \sqrt{47+2}\amp\stackrel{?}{=}7\\ \sqrt{49}\amp\stackrel{\checkmark}{=}7 \end{align*}

   So, the solution set is \(\left\{\frac{47}{3}\right\}\text{.}\)

3. Since the variable \(x\) only appears once, we can apply steps one at a time to undo all of the operations that are done to \(x\) and eventually isolate it.

   \begin{align*} 3(5x-6)-7\amp=2\\ 3(5x-6)\amp=9\\ 5x-6\amp=3\\ 5x\amp=9\\ x\amp=\frac{9}{5} \end{align*}

   The solution set is \(\left\{\frac{9}{5}\right\}\text{.}\)

1. To solve the equation \((x-4)^2+2x=0\text{,}\) note that it is a quadratic equation, and we can write it in standard form.

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

   Now we may use the quadratic formula 7.2.2.

   \begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(16)}}{2(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{\firsthighlight{4}\cdot\secondhighlight{-1}\cdot3}}{2}\\ \amp=\frac{6\pm\sqrt{\firsthighlight{4}}\cdot\sqrt{\secondhighlight{-1}}\cdot\sqrt{3}}{2}\\ \amp=\frac{6\pm\firsthighlight{2}\secondhighlight{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 \(\left\{3-i\sqrt{3},3+i\sqrt{3}\right\}\text{.}\)

2. To solve the equation \(16x-2(3x-1)=3\) we first we first note that it is linear. Since it is linear, we just need to follow the steps outlined in Process 2.1.4.

   \begin{align*} 16x-2(3x-1)\amp=7\\ 16x-6x+3\amp=7\\ 10x+3\amp=7\\ 10x\amp=4\\ x\amp=\frac{4}{10}\\ x\amp=\frac{2}{5} \end{align*}

   So, the solution set is \(\left\{\frac{2}{5}\right\}\text{.}\)

3. Since the equation \(\sqrt{x+2}=x-4\) is a radical equation, we should isolate the radical (which it already is) and square both sides of the equation.

   \begin{align*} \sqrt{x+2}\amp=x-4\\ \left(\sqrt{x+2}\right)^2\amp=\left(x-4\right)^2\\ x+2\amp=x^2-8x+16\\ 0\amp=x^2-9x+14 \end{align*}

   Since the equation is now quadratic, we may use the quadratic formula 7.2.2 to solve it.

   \begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(-9)\pm\sqrt{(-9)^2-4(1)(14)}}{2(1)}\\ \amp=\frac{9\pm\sqrt{81-56}}{2}\\ \amp=\frac{9\pm\sqrt{25}}{2}\\ \amp=\frac{9\pm5}{2} \end{align*}
   \begin{align*} x\amp=\frac{9-5}{2}\amp\amp\text{or}\amp x\amp=\frac{9+5}{2}\\ x\amp=\frac{4}{2}\amp\amp\text{or}\amp x\amp=\frac{14}{2}\\ x\amp=2\amp\amp\text{or}\amp x\amp=7 \end{align*}

   Since this is a radical equation, we should verify our solutions and look out for “extraneous solutions”.

   \begin{align*} \sqrt{\substitute{2}+2}\amp\stackrel{?}{=}\substitute{2}-4\amp\amp\text{or}\amp\sqrt{\substitute{7}+2}\amp\stackrel{?}{=}\substitute{7}-4\\ \sqrt{4}\amp\stackrel{\text{no}}{=}-2\amp\amp\text{or}\amp\sqrt{9}\amp\stackrel{\text{\checkmark}}{=}3 \end{align*}

   So the solution set is \(\{7\}\text{.}\)

1. Since \(x\) only appears once in the euqation, we only need to undo the operations that are done to it.

   \begin{align*} v^2\amp=v_0^2+2ax\\ v^2-v_0^2\amp=2ax\\ \frac{v^2-v_0^2}{2a}\amp=x \end{align*}

   So we find \(x=\frac{v^2-v_0^2}{2a}\text{.}\)

2. Since \(v\) only appears once in the euqation, we only need to undo the operations that are done to it. According to the order of operations, on the right side of the equation,

   1. \(v\) is squared.

   2. The result is negated.

   3. The result is added to \(c^2\text{.}\)

   4. The result has a square root applied.

   5. The result is multiplied by \(\ell_0\text{.}\)

   So we do all of the opposite things in the opposite order.

   \begin{align*} c \ell\amp=\ell_0\sqrt{c^2-v^2}\\ \frac{c\cdot \ell}{\ell_0}\amp=\sqrt{c^2-v^2}\\ \left(\frac{c\cdot \ell}{\ell_0}\right)^2\amp=\left(\sqrt{c^2-v^2}\right)^2\\ \left(\frac{c\cdot \ell}{\ell_0}\right)^2\amp=c^2-v^2\\ \left(\frac{c\cdot \ell}{\ell_0}\right)^2-c^2\amp=-v^2\\ -\left(\frac{c\cdot \ell}{\ell_0}\right)^2+c^2\amp=v^2\\ \pm\sqrt{-\left(\frac{c\cdot \ell}{\ell_0}\right)^2+c^2}\amp=v\\ \pm\sqrt{c^2-\left(\frac{c\cdot \ell}{\ell_0}\right)^2}\amp=v \end{align*}

   So, we find \(v=\pm\sqrt{c^2-\left(\frac{c\cdot \ell}{\ell_0}\right)^2}\text{.}\)

3. This is a quadratic equtaion when we view \(t\) as the variable. First, we should rearrange the equation to standard form.

   \begin{align*} y\amp=\frac{\alpha t^2}{2}+vt\\ 0\amp=\frac{\alpha }{2}t^2+vt-y \end{align*}

   It is helpful with many equations to “clear denominators”. In this case, that means multiplying each side of the equation by \(2\text{.}\)

   \begin{align*} 0\amp=\alpha t^2+2vt-2y \end{align*}

   Now, we may apply the quadratic formula 7.2.2.

   \begin{align*} t\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ t\amp=\frac{-2v\pm\sqrt{(2v)^2-4\alpha (-2y)}}{2\alpha}\\ t\amp=\frac{-2v\pm\sqrt{4v^2+8\alpha y}}{2\alpha}\\ t\amp=\frac{-2v\pm\sqrt{4\left(v^2+2\alpha y\right)}}{2\alpha}\\ t\amp=\frac{-2v\pm2\sqrt{v^2+2\alpha y}}{2\alpha}\\ t\amp=\frac{-v\pm \sqrt{v^2+2\alpha y}}{\alpha} \end{align*}

   So we find \(t=\frac{-v\pm\sqrt{v^2+2\alpha y}}{\alpha}\text{.}\)