11.1A Solving Quadratic Equations Using the Square Root Property
Square Root Property
If 𝑢2 = 𝑑, then 𝑢 = √𝑑 or 𝑢 = −√𝑑.
Solve 𝑥 2 = 9 using the Square Root Property
Example:
By the Square Root Property,
since 𝑥 2 = 9, 𝑥 = √9 = 3 or 𝑥 = −√9 = −3.
By substitution, we can see that both 3 and –3 are solutions:
32 = 9
(−3)2 = 9
Algebraically, the property follows from taking the square root of both sides
of the equation: 𝑢2 = 𝑑
√𝑢2 = √𝑑 or √𝑢2 = −√𝑑
𝑢 = √𝑑 or 𝑢 = −√𝑑
𝑢 = ±√𝑑 ± √𝑑 is a combined form for " √𝑑 or − √𝑑"
NOTE: To be able to use the Square Root Property, the equation must have a
quadratic term and a constant but not a linear term in the variable.
To solve a quadratic equation using the Square Root Property
 isolate the quadratic (squared) term
 divide by the coefficient of the quadratic term
 apply the Square Root Property
 simplify all answers (give exact answers – no decimal approximations)
Solve 4𝑥 2 − 9 = 0 using the Square Root Property
Example:
4𝑥 2 = 9
𝑥 = ±√
𝑥2 =
9
4
𝑥=±
4
Solutions are
9
3
2
3
and − .
2
3
2
Example: Solve 2𝑥 2 + 4 = 4 using the Square Root Property
2𝑥 2 = 0
𝑥2 = 0
𝑥 = √0
𝑥=0
Solution is 0.
Note: zero has only one square root, zero.
-------------------------------------------------------------------------------Example: Solve 𝑥 2 + 6 = 1 using the Square Root Property
𝑥 2 = −5
𝑥 = ±√−5
𝑥 = ±𝑖√5
Solutions are 𝑖√5 and − 𝑖√5 .
-------------------------------------------------------------------------------Example: Solve (𝑥 + 2)2 = 9 using the Square Root Property
(𝑥 + 2)2 is the quadratic term.
𝑥 + 2 = ±√9
𝑥 + 2 = ±3
𝑥 = −2 ± 3
𝑥 = −2 + 3 = 1
𝑥 = −2 − 3 = −5
Solutions are 1 and –5.
-------------------------------------------------------------------------------Example: Solve (𝑥 + 3)2 − 3 = 4 using the Square Root Property
(𝑥 + 3)2 is the quadratic term.
(𝑥 + 3)2 = 7
𝑥 + 3 = ±√7
𝑥 = −3 ± √7
Solutions are −3 + √7 and − 3 − √7 .
Example: Solve 𝑥 2 − 4𝑥 + 4 = 6 using the Square Root Property
This equation has a quadratic and a linear term in x. In this form, it cannot be
solved using the Square Root Property.
But, 𝑥 2 − 4𝑥 + 4 is a perfect − square trinomial and can be written as
(𝑥 − 2)2
The original equation can now be written as (𝑥 − 2)2 = 6 and the Square
Root Property can be used.
𝑥 − 2 = ±√6
𝑥 = 2 ± √6
Solutions are 2 + √6 and 2 − √6
-------------------------------------------------------------------------------Example: Solve 3𝑥 2 − 16 = 0 using the Square Root Property
3𝑥 2 = 16
𝑥 = ±√
𝑥=±
𝑥=±
𝑥2 =
16
16
3
𝑥=±
3
4
𝑥=±
√3
√16
√3
4∙√3
√3∙√3
4√3
3
4√3
4√3
Solutions are
and −
3
3
-------------------------------------------------------------------------------Solving Quadratic Equations by Completing the Square
Forming Perfect-square Trinomials
A perfect–square trinomial with a leading coefficient of 1 is the square of a
binomial, 𝑥 + 𝑏.
(𝑥 + 𝑏)2 = 𝑥 2 + 2𝑏𝑥 + 𝑏 2
To form a perfect–square trinomial, given the quadratic and linear terms 𝑥 2 +
2𝑏𝑥 , do the following:
1. Find ½ of the coefficient of the linear term, 2b, OR divide the
coefficient of the linear term by two. The result is the second term
of the binomial being squared, b.
1
2𝑏
∙ 2𝑏 = 𝑏
=𝑏
2
2
2. Square the result from 1. This is b2, the third term of the trinomial.
3. Add b2 to the other two terms to form the perfect–square trinomial.
x2 + 2bx + b2
Once you have the perfect–square trinomial, it will be factored and written as
(𝑥 + 𝑏)2 .
Example: 𝑥 2 + 6𝑥
2𝑏 = 6
2𝑏
2
6
=2
𝑏 2 = 32 = 9
𝑏=3
Perfect-square trinomial is 𝑥 2 + 6𝑥 + 9
Factored form of the trinomial is (𝑥 + 3)2
-------------------------------------------------------------------------------Example: 𝑥 2 − 3𝑥
2𝑏 = −3
2𝑏
2
=
−3
2
𝑏=−
Perfect-square trinomial is 𝑥 2 − 3𝑥 +
3
2
3 2
9
2
4
𝑏 2 = (− ) =
9
4
3 2
Factored form of the trinomial is (𝑥 − )
2
Note: b is negative.
Perfect–Square Trinomials and Completing the Square
Now that we are able to form a perfect–square trinomial, we are ready to
solve quadratic equations by completing the square.
Solve x2 + 6x – 1 = 0
NOTE: This equation cannot be solved
by factoring.
1. Add 1 to both sides of the equation. This leaves only the quadratic
and linear terms on the left side of the equation.
x2 + 6x – 1 + 1 = 0 + 1
x2 + 6x = 1
2. Find the third term for the perfect–square trinomial (see Example).
Add this to both sides of the equation.
x2 + 6x + 9 = 1 + 9
3. Factor the perfect–square trinomial as the square of a binomial.
Simplify the expression on the right side.
(𝑥 + 3)2 = 10
4. Apply the Square Root Property to solve the equation.
𝑥 + 3 = ±√10
𝑥 = −3 ± √10
Solutions are −3 + √10 and − 3 − √10
--------------------------------------------------------------------------------------------Leading Coefficient of the Quadratic Term is Not 1
Start by dividing both sides of the equation by the leading coefficient and
then solve as before.
Solve 2x2 + 12x – 6 = 0
1.
2.
2𝑥 2 +12𝑥−6
2
=
0
2
𝑥 2 + 6𝑥 − 3 = 0
NOTE: This equation cannot be solved
by factoring.
3. 𝑥 2 + 6𝑥 = 3
4. 𝑥 2 + 6𝑥 + 9 = 3 + 9
5. (𝑥 + 3)2 = 12
𝑥 + 3 = ±√12
√12 = √4 ∙ 3 = √4 ∙ √3 = 2√3
𝑥 + 3 = ±2√3
𝑥 = −3 ± 2√3
Solutions are −3 + 2√3 and − 3 − 2√3