10.4: Use Square Roots to Solve Quadratic Equations
What if you had a quadratic equation like
Watch This
CK12 Foundation: 1004S Solving Quadratic Equations Using Square Roots
Guidance
So far you know how to solve quadratic equations by factoring. However, this method works only if a quadratic polynomial can be factored. In the real world, most quadratics can’t be factored, so now we’ll start to learn other methods we can use to solve them. In this Concept, we’ll examine equations in which we can take the square root of both sides of the equation in order to arrive at the result.
Solve Quadratic Equations Involving Perfect Squares
Let’s first examine quadratic equations of the type
We can solve this equation by isolating the
Once the
Often this is written as
Example A
Solve the following quadratic equations:
a)
b)
Solution
a)
Isolate the
Take the square root of both sides:
The solutions are
b)
Isolate the
Take the square root of both sides:
The solutions are
We can also find the solution using the square root when the
We just have to isolate the
Then we can take the square root of both sides of the equation:
Often this is written as:
Example B
Solve the following quadratic equations.
a)
b)
Solution
a)
Isolate the
Take the square root of both sides:
Answer:
b)
Isolate the
Answer:
As you’ve seen previously, some quadratic equations have no real solutions.
Example C
Solve the following quadratic equations.
a)
b)
Solution
a)
Isolate the
Take the square root of both sides: \begin{align*} x = \sqrt{1}\end{align*}
Square roots of negative numbers do not give real number results, so there are no real solutions to this equation.
b) \begin{align*} 4x^2 + 9 = 0\end{align*}
Isolate the \begin{align*}x^2\end{align*}
\begin{align*}4x^2 & = 9\\ x^2 & =  \frac{9}{4}\end{align*}
There are no real solutions.
We can also use the square root function in some quadratic equations where both sides of an equation are perfect squares. This is true if an equation is of this form:
\begin{align*}(x  2)^2 = 9\end{align*}
Both sides of the equation are perfect squares. We take the square root of both sides and end up with two equations: \begin{align*}x  2 = 3\end{align*}
Solving both equations gives us \begin{align*}x = 5\end{align*}
Example D
Solve the following quadratic equations.
a) \begin{align*}(x  1)^2 = 4\end{align*}
b) \begin{align*}(x + 3)^2 = 1\end{align*}
Solution
a) \begin{align*}(x  1)^2 = 4\end{align*}
\begin{align*}\text{Take the square root of both sides}: & & x  1 & = 2 \ \text{and} \ x  1 = 2\\ \text{Solve each equation}: & & x & = 3 \ \text{and} \ x = 1\end{align*}
Answer: \begin{align*}x =3\end{align*}
b) \begin{align*}(x + 3)^2 = 1\end{align*}
\begin{align*}\text{Take the square root of both sides}: & & x + 3 & = 1 \ \text{and} \ x + 3 = 1\\ \text{Solve each equation}: & & x & = 2 \ \text{and} \ x = 4\end{align*}
Answer: \begin{align*}x = 2\end{align*}
It might be necessary to factor the righthand side of the equation as a perfect square before applying the method outlined above.
Watch this video for help with the Examples above.
CK12 Foundation: 1004 Solving Quadratic Equations Using Square Roots
Vocabulary
 The solutions of a quadratic equation are often called the roots or zeros.
Guided Practice
Solve the following quadratic equations.
a) \begin{align*}x^2 + 8x + 16 = 25\end{align*}
b) \begin{align*}4x^2  40x + 25 = 9\end{align*}
Solution
a) \begin{align*}x^2 + 8x + 16 = 25\end{align*}
\begin{align*}& \text{Factor the righthandside}: & & x^2 + 8x + 16 = (x + 4)^2 \quad \text{so} \quad (x + 4)^2 = 25\\ & \text{Take the square root of both sides}: & & x + 4 = 5 \ \text{and} \ x + 4 = 5 \\ & \text{Solve each equation}: & & x = 1 \ \text{and} \ x = 9 \end{align*}
Answer: \begin{align*}x = 1\end{align*}
b) \begin{align*}4x^2  20x + 25 = 9\end{align*}
\begin{align*}& \text{Factor the righthandside}: & & 4x^2  20x + 25 = (2x  5)^2 \quad \text{so} \quad (2x  5)^2 = 9\\ & \text{Take the square root of both sides}: & & 2x  5 = 3 \ \text{and} \ 2x  5 = 3 \\ & \text{Solve each equation}: & & 2x = 8 \ \text{and} \ 2x = 2 \end{align*}
Answer: \begin{align*}x = 4\end{align*}
Practice
Solve the following quadratic equations.

\begin{align*}x^2  1 = 0\end{align*}
x2−1=0 
\begin{align*}x^2  100 = 0\end{align*}
x2−100=0 
\begin{align*}x^2 + 16 = 0\end{align*}
x2+16=0 
\begin{align*}9x^2  1 = 0\end{align*}
9x2−1=0 
\begin{align*}4x^2  49 = 0\end{align*}
4x2−49=0 
\begin{align*}64x^2  9 = 0\end{align*}
64x2−9=0 
\begin{align*}x^2  81 = 0\end{align*}
x2−81=0 
\begin{align*}25x^2  36 = 0\end{align*}
25x2−36=0 
\begin{align*}x^2 + 9 = 0\end{align*}
x2+9=0 
\begin{align*}x^2  16 = 0\end{align*}
x2−16=0 
\begin{align*}x^2  36 = 0\end{align*}
x2−36=0 
\begin{align*}16x^2  49 = 0\end{align*}
16x2−49=0 
\begin{align*}(x  2)^2 = 1\end{align*}
(x−2)2=1 
\begin{align*}(x + 5)^2 = 16\end{align*}
(x+5)2=16 
\begin{align*}(2x  1)^2  4 = 0\end{align*}
(2x−1)2−4=0 
\begin{align*}(3x + 4)^2 = 9\end{align*}
(3x+4)2=9 
\begin{align*}(x  3)^2 + 25 = 0\end{align*}
(x−3)2+25=0  \begin{align*}x^2  10x + 25 =9\end{align*}
 \begin{align*}x^2 + 18x + 81 = 1\end{align*}
 \begin{align*}4x^2  12x + 9 = 16\end{align*}
 \begin{align*}2(x + 3)^2 = 8\end{align*}
Image Attributions
Description
Learning Objectives
Here you'll learn how to solve quadratic equations in which finding the solutions involves square roots.