What if a bird were flying through the air carrying a worm in its beak? All of the sudden, the worm slips from the bird's beak, and the poor bird no longer has his dinner! If the bird were flying at a height of 2000 feet, how long would it take the worm to hit the ground, assuming there were no air resistance? In this Concept, you'll learn to solve quadratic equations like the one representing this scenario by using square roots.

### Guidance

Suppose you needed to find the value of \begin{align*}x\end{align*}

- Make a table of values.
- Graph this equation as a system.
- Cancel the square using its inverse operation.

The inverse of a square is a square root.

By applying the square root to each side of the equation, you get:

\begin{align*}x &= \pm \sqrt{81}\\
x &= 9 \ or \ x=-9\end{align*}

In general, the solution to a quadratic equation of the form \begin{align*}0=ax^2-c\end{align*}

\begin{align*}x=\sqrt{\frac{c}{a}} \ \text{or} \ x=- \sqrt{\frac{c}{a}}\end{align*}

Another way to write this is \begin{align*}x=\pm \sqrt{\frac{c}{a}}.\end{align*}

#### Example A

*Solve for x when \begin{align*}0=2x^2-18\end{align*}.*

**Solution:**

\begin{align*}0=2x^2-18 \Rightarrow 18=2x^2\Rightarrow 9=x^2 \Rightarrow x=\pm 3.\end{align*}

This is the same result you would get if you use the formula above:

\begin{align*}x=\pm \sqrt{\frac{c}{a}}=\pm \sqrt{\frac{18}{2}}=\pm \sqrt{9}=\pm 3.\end{align*}

#### Example B

*Solve* \begin{align*}(x-4)^2-9=0\end{align*}.

**Solution:**

Begin by adding 9 to each side of the equation.

\begin{align*}(x-4)^2=9\end{align*}

Cancel the square by taking the square root.

\begin{align*}x-4=3 \ or \ x-4=-3\end{align*}

Solve both equations for \begin{align*}x: x=7 \ or \ x=1\end{align*}

**Real-World Problems Using Square Roots**

In the previous Concept, you learned Newton’s formula for projectile motion. Let’s examine a situation in which there is no initial velocity. When a ball is dropped, there is no outward force placed on its path; therefore, there is no initial velocity.

#### Example C

*A ball is dropped from a 40-foot building. When does the ball reach the ground?*

**Solution:**

Using the equation from the previous Concept, \begin{align*}h(t)=-\frac{1}{2} (g) t^2+v_0 t+h_0\end{align*}, and substituting the appropriate information, you get:

\begin{align*}&& 0 &=-\frac{1}{2} (32)t^2+(0)t+40\\ \text{Simplify:} && 0 &= -16t^2+40\\ \text{Solve for} \ x: && -40 &= -16t^2\\ && 2.5 &= t^2\\ && t & \approx 1.58 \ and \ t \approx -1.58\end{align*}

Because \begin{align*}t\end{align*} is in seconds, it does not make much sense for the answer to be negative. So the ball will reach the ground at approximately 1.58 seconds.

### Guided Practice

*A rock is dropped from the top of a cliff and strikes the ground 7.2 seconds later. How high is the cliff in meters?*

**Solution:**

Using Newton’s formula, substitute the appropriate information.

\begin{align*}&& 0 &= -\frac{1}{2} (9.8)(7.2)^2+(0)(7.2)+ h_0\\ \text{Simplify:} && 0 &= -254.016+h_0\\ \text{Solve for} \ h_0: && h_0 &= 254.016\end{align*}

The cliff is approximately 254 meters tall.

### Practice

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Solving Quadratic Equations by Square Roots (11:03)

Solve each quadratic equation.

- \begin{align*}x^2=196\end{align*}
- \begin{align*}x^2-1=0\end{align*}
- \begin{align*}x^2-100=0\end{align*}
- \begin{align*}x^2+16=0\end{align*}
- \begin{align*}9x^2-1=0\end{align*}
- \begin{align*}4x^2-49=0\end{align*}
- \begin{align*}64x^2-9=0\end{align*}
- \begin{align*}x^2-81=0\end{align*}
- \begin{align*}25x^2-36=0\end{align*}
- \begin{align*}x^2+9=0\end{align*}
- \begin{align*}x^2-16=0\end{align*}
- \begin{align*}x^2-36=0\end{align*}
- \begin{align*}16x^2-49=0\end{align*}
- \begin{align*}(x-2)^2=1\end{align*}
- \begin{align*}(x+5)^2=16\end{align*}
- \begin{align*}(2x-1)^2-4=0\end{align*}
- \begin{align*}(3x+4)^2=9\end{align*}
- \begin{align*}(x-3)^2+25=0\end{align*}
- \begin{align*}x^2-6=0\end{align*}
- \begin{align*}x^2-20=0\end{align*}
- \begin{align*}3x^2+14=0\end{align*}
- \begin{align*}(x-6)^2=5\end{align*}
- \begin{align*}(4x+1)^2-8=0\end{align*}
- \begin{align*}(x+10)^2=2\end{align*}
- \begin{align*}2(x+3)^2=8\end{align*}
- How long does it take a ball to fall from a roof to the ground 25 feet below?
- Susan drops her camera in the river from a bridge that is 400 feet high. How long is it before she hears the splash?
- It takes a rock 5.3 seconds to splash in the water when it is dropped from the top of a cliff. How high is the cliff in meters?
- Nisha drops a rock from the roof of a building 50 feet high. Ashaan drops a quarter from the top-story window, which is 40 feet high, exactly half a second after Nisha drops the rock. Which hits the ground first?
- Victor drops an apple out of a window on the \begin{align*}10^{th}\end{align*} floor, which is 120 feet above ground. One second later, Juan drops an orange out of a \begin{align*}6^{th}\end{align*}-floor window, which is 72 feet above the ground. Which fruit reaches the ground first? What is the time difference between the fruits’ arrival to the ground?

**Mixed Review**

- Graph \begin{align*}y=2x^2+6x+4\end{align*}. Identify the following:
- Vertex
- \begin{align*}x-\end{align*}intercepts
- \begin{align*}y-\end{align*}intercepts
- axis of symmetry

- What is the difference between \begin{align*}y=x+3\end{align*} and \begin{align*}y=x^2+3\end{align*}?
- Determine the domain and range of \begin{align*}y=-(x-2)^2+7\end{align*}.
- The Glee Club is selling hot dogs and sodas for a fundraiser. On Friday the club sold 112 hot dogs and 70 sodas and made $154.00. On Saturday the club sold 240 hot dogs and 120 sodas and made $300.00. How much is each soda? Each hot dog?