Sometimes, a quadratic equation has a radical of 2, but this does not imply that such quadratic equations are not factorisable. The following example shows how you can split a quadratic equation having a √2 (or a square root of 2) in it.
Quadratic equation: 2x2 + √2x - 2 = 0
Quadratic equation: 2x2 + √2x - 2 = 0
- Master product (product of first and last terms' coefficients) = 2 * -2 = -4
- Coefficient of middle term = √2
Now you have to find two numbers which add up to give √2 and whose product is -4. Thus
From the above example, it is clear that when there is a √2 in a quadratic equation, it can be split by considering the radical √2 as a variable itself. As you do not pay attention to the variable x when trying to split the middle term of a quadratic equation, similarly a radical of √2 is also considered a variable. However, the radical will be considered when calculating the master product. The two parts obtained after splitting the middle term should each contain a radical of √2.
__ + __ = √2
__ * __ = -4Try to fill in the above blanks with two numbers, each of which contains a √2 in it. The two numbers are:
2√2 + -√2 = √2
2√2 * -√2 = 4Thus, the two numbers are 2√2 and -√2. Thus the quadratic equation's middle term can be split as follows:
2x2 + √2x - 2 = 0
2x2 + 2√2x + -√2x - 2 = 0Factoring the quadratic equation further you obtain:
2x(x + √2) - √2(x - √2) = 0
(x + √2) (2x - √2) = 0Thus the quadratic equation 2x2 + √2x - 2 = 0 has been factorized into (x + √2) (2x - √2) = 0.
From the above example, it is clear that when there is a √2 in a quadratic equation, it can be split by considering the radical √2 as a variable itself. As you do not pay attention to the variable x when trying to split the middle term of a quadratic equation, similarly a radical of √2 is also considered a variable. However, the radical will be considered when calculating the master product. The two parts obtained after splitting the middle term should each contain a radical of √2.
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