The Difference of Perfect Squares
QUADRATIC FUNCTION
ax^{2} + bx + c = 0
Where a, b, and c are constants and a ≠ 0
Some examples are:
x^{2} + 3x  3 = 0 (Where a = 1, b = 3 and c = 3)
4x^{2} + 9 = 0 (Where a = 4, b = 0 and c = 9)
x^{2} + 5x = 0 (Where a = 1, b = 5 and c = 0)
One way to solve quadratic equations is to factor the equation, set each factor to zero and solve for x.
This is the topic of "Factoring Quadratics when a = 1" and "Factoring Quadratic Equations when a ≠ 1".
Factoring can be made simpler when the quadratic equation can be written as a difference of perfect squares.
DIFFERENCE OF PERFECT SQUARES
p^{2}  r^{2} = 0 → (p + r)(p  r)
A quadratic equation can be written as the difference of perfect squares when the coefficients, the numbers multiplied by the variables, are perfect squares and the exponent of each term is even.
Let's solve the following using this factoring rule:
Step 1: Write the equation in the general form ax^{2} + bx + c = 0. This equation is already in the proper form. 
x^{2}  36 = 0 
Step 2: Write the equation as a difference of perfect squares a^{2}  b^{2} = 0. x^{2} is already written as a perfect square and 36 written as a perfect square is 6^{2}. 
(x)^{2}  6^{2} = 0 
Step 3: Factor the difference of perfect squares by applying the factor rule. p^{2}  r^{2} = (p + r)(p  r) 
p = x, r = 6 (x + 6)(x  6) = 0 
Step 4: Set each factor to zero and solve for x.

(x + 6) = 0, or (x  6) = 0 x = 6, or x = 6 
Step 5: Check each answer 
x = 6: 6^{2}  6^{2} = 0 0 = 0 x = 6: (6)^{2}  6^{2} = 0 0 = 0 
Step 1: Write the equation in the general form ax^{2} + bx + c = 0. Add 37 to both sides. 
4x^{2}  25 = 0 
Step 2: Write the equation as a difference of perfect squares p^{2}  r^{2} = 0. 4x^{2} written as a perfect square is (2x)^{2} and 25 written as a perfect square is 5^{2}, therefore p = 2x, r = 5. 
(2x)^{2}  5^{2} = 0 
Step 3: Factor the difference of perfect squares by applying the factor rule. p^{2}  r^{2} = (p + r)(p  r) 
p = 2x, r = 5 (2x + 5)(2x  5) = 0 
Step 4: Set each factor to zero and solve for x. 
(2x + 5) = 0, or (2x  5) = 0 $x=\frac{5}{2}$, or $x=\frac{5}{2}$ 
Step 5: Check each answer 
$x=\frac{5}{2}:$ $4{\left(\frac{5}{2}\right)}^{2}25=0$ 4(6.25)  25 = 0 25  25 = 0 0 = 0 $x=\frac{5}{2}:$ $4{\left(\frac{5}{2}\right)}^{2}25=0$ 4(6.25)  25 = 0 25  25 = 0 0 = 0 
Step 1: Write the equation in the general form ax^{2} + bx + c = 0. This equation is already in the proper form. 
49x^{4}  64x^{2} = 0 
Step 2: Notice that you can factor out a x^{2} from the equation. Always look for opportunities to simplify the expression before applying the perfect square rule. 
x^{2}(49x^{2}  64) = 0 
Step 3: Write the parenthetical expression as a difference of perfect squares p^{2}  r^{2} = 0 49x^{2} written as a perfect square is (7x)^{2} and 64 written as a perfect square is 8^{2}, therefore p = 7x, r = 8. 
x^{2}{(7x)^{2}  (8)^{2}} = 0 
Step 4: Factor the difference of perfect square by applying the factor rule. p^{2}  r^{2} = (p + r)(p  r) 
p = 7x, r = 8 x^{2}(7x + 8)(7x  8) = 0 
Step 5: Set each factor equal to zero and solve for x. 
x^{2} = 0, (7x + 8) = 0, or (7x  8) = 0 $x=0,x=\frac{8}{7}$, or $x=\frac{8}{7}$ 
Step 5: Check each answer 
x = 0 49(0)^{4}  64(0)^{2} = 0 0 = 0 $x=\frac{8}{7}:$ $49{\left(\frac{8}{7}\right)}^{4}64{\left(\frac{8}{7}\right)}^{2}=0$ 0 = 0 $x=\frac{8}{7}:$ $49{\left(\frac{8}{7}\right)}^{4}64{\left(\frac{8}{7}\right)}^{2}=0$ 0 = 0 
Related Links: Math algebra Completing the Square when a equals 1 Introduction and Simple Equations Simple Equations with the Natural Base Algebra Topics 
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