Elementary Algebra

For each equation, we find the square root of the right side. (That is, we find a number whose square gives the right side.) Remember that every positive number has two square roots.

1. Because \(~~100^2=10,000,~~t=\pm 100\)

2. Because \(~~\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9},~~b=\pm \dfrac{2}{3}\)

3. Because \(~~0.2^2=0.04,~~h=\pm 0.2\)

4. Because \(~~17^2=(-17)^2,~~m=\pm 17\)

We first solve for \(x^2\) as follows.

Once we have isolated \(x^2\text{,}\) we take the square root of each side to find

The exact solutions are thus \(\sqrt{6}\) and \(-\sqrt{6}\text{.}\) (Don't forget that every positive number has two square roots.) We can also find decimal approximations for the solutions using a calculator. Rounded to two decimal places, the solutions are \(2.45\) and \(-2.45\text{.}\)

1. In this question, we are given the value of \(t\) and asked to find the corresponding value of \(h\text{.}\) To do this we evaluate the formula for \(t=0.5\text{.}\) We substitute \(\alert{0.5}\) for \(t\) into the formula, and simplify.

   \begin{align*} h \amp = 20-16(\alert{0.5})^2 \amp \amp \blert{\text{Compute the power.}}\\ \amp = 20-16(0.25) \amp \amp \blert{\text{Multiply, then subtract.}}\\ \amp = 20-4=16 \end{align*}

   The cat is 16 feet above the ground after 0.5 second. You can also use your calculator to simplify the expression for by entering

   \begin{equation*} 20~~\boxed{~-~}~~16~~\boxed{~\times~}~~0.5~~\boxed{~x^2~}~~\boxed{\text{ENTER}} \end{equation*}

2. We would like to find the value of \(t\) when the height, \(h\text{,}\) is known. We substitute \(h=0\) into the equation to obtain

   \begin{equation*} \alert{0}=20-16t^2 \end{equation*}

   To solve this equation we use extraction of roots. We first isolate \(t^2\) on one side of the equation.

   \begin{align*} 16t^2 \amp = 20 \amp \amp \blert{\text{Divide by 16.}}\\ t^2 \amp = \dfrac{20}{16} = 1.25 \end{align*}

   Next, we take the square root of both sides of the equation to find

   \begin{equation*} t= \pm \sqrt{1.25} \approx \pm 1.118 \end{equation*}

   Only the positive solution makes sense here, so the cat has approximately 1.12 seconds to be in position for landing. A graph of the cat's height after \(t\) seconds is shown at right. The points corresponding to parts (a) and (b) are labeled.

   

Because the variable we want is squared, we use extraction of roots. First, we multiply both sides by 3 to clear the fraction.

Because the radius of a cone must be a positive number, we use only the positive square root: \(r=\sqrt{\dfrac{3V}{\pi h}}\text{.}\)