Answer:
9.1 Extracting Square Roots
LEARNING OBJECTIVE
Solve quadratic equations by extracting square roots.
Extracting Square Roots
Recall that a quadratic equation is in standard form if it is equal to 0:
where a, b, and c are real numbers and
a
≠
0
. A solution to such an equation is called a root. Quadratic equations can have two real solutions, one real solution, or no real solution. If the quadratic expression on the left factors, then we can solve it by factoring. A review of the steps used to solve by factoring follow:
Step 1: Express the quadratic equation in standard form.
Step 2: Factor the quadratic expression.
Step 3: Apply the zero-product property and set each variable factor equal to 0.
Step 4: Solve the resulting linear equations.
For example, we can solve
x
2
−
4
=
0
by factoring as follows:
The two solutions are −2 and 2. The goal in this section is to develop an alternative method that can be used to easily solve equations where b = 0, giving the form
The equation
x
2
−
4
=
0
is in this form and can be solved by first isolating
x
2
.
If we take the square root of both sides of this equation, we obtain the following:
Here we see that
x
=
−
2
and
x
=
2
are solutions to the resulting equation. In general, this describes the square root property; for any real number k,
The notation “±” is read “plus or minus” and is used as compact notation that indicates two solutions. Hence the statement
x
=
±
√
k
indicates that
x
=
−
√
k
or
x
=
√
k
. Applying the square root property as a means of solving a quadratic equation is called extracting the roots.
Example 1: Solve:
x
2
−
25
=
0
.
Solution: Begin by isolating the square.
Next, apply the square root property.
Answer: The solutions are −5 and 5. The check is left to the reader.
Certainly, the previous example could have been solved just as easily by factoring. However, it demonstrates a technique that can be used to solve equations in this form that do not factor.
Example 2: Solve:
x
2
−
5
=
0
.
Solution: Notice that the quadratic expression on the left does not factor. We can extract the roots if we first isolate the leading term,
x
2
.
Apply the square root property.
For completeness, check that these two real solutions solve the original quadratic equation. Generally, the check is optional.
Answer: The solutions are
−
√
5
and
√
5
.
Example 3: Solve:
4
x
2
−
45
=
0
.
Solution: Begin by isolating
x
2
.
Apply the square root property and then simplify.
Answer: The solutions are
−
3
√
5
2
and
3
√
5
2
.
Sometimes quadratic equations have no real solution.
Example 4: Solve:
x
2
+
9
=
0
.
Solution: Begin by isolating
x
2
.
After applying the square root property, we are left with the square root of a negative number. Therefore, there is no real solution to this equation.
Answer: No real solution
Reverse this process to find equations with given solutions of the form ±k.
Example 5: Find an equation with solutions
−
2
√
3
and
2
√
3
.
Solution: Begin by squaring both sides of the following equation:
Lastly, subtract 12 from both sides and present the equation in standard form.
Answer:
x
2
−
12
=
0
Try this! Solve:
9
x
2
−
8
=
0
.
Answer:
x
=
−
2
√
2
3
or
x
=
2
√
2
3
Video Solution
(click to see video)
Consider solving the following equation:
To solve this equation by factoring, first square
x
+
2
and then put it in standard form, equal to zero, by subtracting 25 from both sides.
Factor and then apply the zero-product property.
The two solutions are −7 and 3.
When an equation is in this form, we can obtain the solutions in fewer steps by extracting the roots.
Example 6: Solve:
(
x
+
2
)
2
=
25
.
Solution: Solve by extracting the roots.
At this point, separate the “plus or minus” into two equations and simplify each individually.
Answer: The solutions are −7 and 3.
In addition to fewer steps, this method allows us to solve equations that do not factor.
Example 7: Solve:
(
3
x
+
3
)
2
−
27
=
0
.
Solution: Begin by isolating the square.
Next, extract the roots and simplify.
Solve for x.
Answer: The solutions are
−
1
−
√
3
and
−
1
+
√
3
.
Example 8: Solve:
9
(
2
x
−
1
)
2
−
8
=
0
.
Solution: Begin by isolating the square factor.
Apply the square root property and solve.
Answer: The solutions are
3
−
2
√
2
6
and
3
+
2
√
2
6
.
Try this! Solve:
3
(
x
−
5
)
2
−
2
=
0
.
Answer:
15
±
√
6
3
Video Solution
(click to see video)
Example 9: The length of a rectangle is twice its width. If the diagonal measures 2 feet, then find the dimensions of the rectangle.
Solution:
The diagonal of any rectangle forms two right triangles. Thus the Pythagorean theorem applies. The sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse:
Solve.
Here we obtain two solutions,
w
=
−
2
√
5
and
w
=
2
√
5
. Since the problem asked for a length of a rectangle, we disregard the negative answer. Furthermore, we will rationalize the denominator and present our solutions without any radicals in the denominator.
Back substitute to find the length.
Answer: The length of the rectangle is
4
√
5
5
feet and the width is
2
√
5
5
feet
