# The Difference of Perfect Squares

A quadratic equation is an equation that contains a squared variable as its highest power on any variable. The general form of a quadratic equation is:

ax2 + bx + c = 0

Where a, b, and c are constants and a ≠ 0

Some examples are:

x2 + 3x - 3 = 0     (Where a = 1, b = 3 and c = -3)

4x2 + 9 = 0          (Where a = 4, b = 0 and c = 9)

x2 + 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

p2 - r2 = 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:

x2 - 36 = 0
 Step 1: Write the equation in the general form ax2 + bx + c = 0. This equation is already in the proper form. x2 - 36 = 0 Step 2: Write the equation as a difference of perfect squares a2 - b2 = 0. x2 is already written as a perfect square and 36 written as a perfect square is 62. (x)2 - 62 = 0 Step 3: Factor the difference of perfect squares by applying the factor rule. p2 - r2 = (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: 62 - 62 = 0 0 = 0 x = -6: (-6)2 - 62 = 0 0 = 0
Example 1:      4x2 + 12 = -37
 Step 1: Write the equation in the general form ax2 + bx + c = 0. Add 37 to both sides. 4x2 - 25 = 0 Step 2: Write the equation as a difference of perfect squares p2 - r2 = 0. 4x2 written as a perfect square is (2x)2 and 25 written as a perfect square is 52, therefore p = 2x, r = 5. (2x)2 - 52 = 0 Step 3: Factor the difference of perfect squares by applying the factor rule. p2 - r2 = (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
Example 2:      49x4 - 64x2 = 0
 Step 1: Write the equation in the general form ax2 + bx + c = 0. This equation is already in the proper form. 49x4 - 64x2 = 0 Step 2: Notice that you can factor out a x2 from the equation. Always look for opportunities to simplify the expression before applying the perfect square rule. x2(49x2 - 64) = 0 Step 3: Write the parenthetical expression as a difference of perfect squares p2 - r2 = 0 49x2 written as a perfect square is (7x)2 and 64 written as a perfect square is 82, therefore p = 7x, r = 8. x2{(7x)2 - (8)2} = 0 Step 4: Factor the difference of perfect square by applying the factor rule. p2 - r2 = (p + r)(p - r) p = 7x, r = 8 x2(7x + 8)(7x - 8) = 0 Step 5: Set each factor equal to zero and solve for x. x2 = 0, (7x + 8) = 0, or (7x - 8) = 0 , 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
Factoring quadratics is generally the easier method for solving quadratic equations. However,not all quadratic equations can be factored evenly or are perfect squares. In these cases it is usually better to solve by completing the square or using the quadratic formula.

 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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