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All quadratic equations in one variable can be solved using the method of Completing the Square. This method depends upon the fact:

(a + b)2 = a2 + 2ab + b2

The middle term is twice the product of the square roots of the other two terms:

2ab = 2√a2√b2

Factor 5x2 − 30 = x2 − 5

5x2 − x2 − 30 + 5 = 0

4x2 − 25 = 0

x2 = 25/4

x = ±√(25/4) = ±5/2

Factor the difficult equation,

2x2 − 7x = x + 14

Place all math terms on one side of the equation.

2x2 − x − 7x − 14 = 0

2x2 − 8x − 14 = 0

x2 − 4x − 7 = 0

There is no number combination of the constant 14, [ 1, 14 ] and [ 2, 7 ], that when factored using the “(x )(x )” setup are equivalent to the original equation.

However, x2 − 4x of the quadratic equation is almost a perfect square. The perfect trinomial square x2 − 4x + 4 has factors:

(x − 2) (x − 2).

(x − 2) (x − 2) = x2 − 4x + 4

Isolate the almost perfect square to one side of the equation (very important math step).

x2 − 4x − 7 = 0

x2 − 4x = 7

Use the math Substitution Property.

x2 − 4x = 7

(x − 2) (x − 2) = 7 + 4

(x − 2)2 = 11

x − 2 = ±√11

x = 2 ±√11

To determine the missing term of x2 + 6x to complete a perfect trinomial square:

Apply the form (a + b)2 = a2 + 2ab + b2,

x2 + 6x = a2 + 6a, and

2ab = 6a, then

2b = 6

b = 3, and

(a + 3)2 = a2 + 2a3 + 32, and

(a + 3) (a + 3) = a2 + 6a + 9, when a = 3

Check the equation is correct for values of a = 3 and b = 3

(a + b)2 = a2 + 2ab + b2

(3 + 3)2 = 32 + (2) (3) (3) + 32

62 = 9 + 18 + 9

36 = 36 is correct.

The perfect trinomial math square is:

x2 + 6x + 9 = 36, can be factored as

(x + 3)2 = (3 + 3) (3 + 3) = 36 when x = 3,

and equals 0 when x = −3