Sometimes, a quadratic equation has a radical of

Quadratic equation:

*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:

*2x*^{2}+ √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

From the above example, it is clear that when there is a

*√2*and whose product is*-4*. Thus__ + __ =√2

__ * __ =Try to fill in the above blanks with two numbers, each of which contains a-4

*√2*in it. The two numbers are:2√2 + -√2 =√2

Thus, the two numbers are2√2 * -√2 = 4

*2√2*and*-√2*. Thus the quadratic equation's middle term can be split as follows:2x^{2}+ √2x - 2 = 0

Factoring the quadratic equation further you obtain:2x^{2}+2√2x + -√2x - 2 = 0

2x(x^{}+√2) -√2(x -√2)= 0

Thus the quadratic equation(x +√2)(2x -√2)= 0

*2x*has been factorized into^{2}+ √2x - 2 = 0*(x +**√2)**(2**x -**√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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