# 10.3: Solving Quadratic Equations Using Square Roots

**At Grade**Created by: CK-12

Suppose you needed to find the value of \begin{align*}x\end{align*} such that \begin{align*}x^2=81\end{align*}. How could you solve this equation?

- 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*} is:

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

**Example 1:** *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 square by applying 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*}

In the previous lesson, 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.

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

Using the equation from the previous lesson, \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.

Example: *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 Set

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

- How are the graphs of \begin{align*}y=x+3\end{align*} and \begin{align*}y=x^2+3\end{align*} different?
- 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. For how much did the Glee Club sell each soda and each hot dog?

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