X 2 36

Math problem: Factor x

you are right. These factors are important.

(x + 6i) (x6i)

X 2 36

36X2
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Math Problem: Factor x 2 + 36?
I have a math problem. The problem is to factor x x 2 + 36 and the question is whether you use real or imaginary numbers. I suggest solutions without solutions and without imaginary numbers. Any suggestions on this?
36x 2 65x36 If all else fails then apply the square formula and go back from there x = (65 + / sqrt (65 2 4 * 36 * (36)) / (2 * 36) x = ( 65 + / square (65 2 + 72 2)) / 72 x = (65 + / square (97 2)) / 72 x = (65 + / 97) / 72 x = 162/72, 32/72 x = 81/36, 4/9 x = 9/4, 4/9 (4x 9) * (9x + 4) 9s 2 6. + t 2 (3s t) 2 3 * (x 2 + 4) 4 * 2x * (x 3) 4 + (x 2 + 4) 5 * (x 3) 3 => (x 2 + 4) 4 * (x 3) 3 * ( 3 * 2x * (x 3) + (x 2 + 4)) => (x 2 + 4) 4 * (x3) 3 * (6x * (x3) + x 2 + 4) => (x 2) +4) 4 * (x3) 3 * (6x 2 18x + x 2 + 4) => (x 2 + 4) 4 * (x 3) 3 * (7x 2 18x + 4))
The answer is an imaginary number. This is the only way to consider it.
(x 6i) (x + 6i) You can change this by inserting = into z and subtracting 36 from both sides to solve the equation.
x 2 = 36 And this is not possible with real numbers. So they are odd numbers.

X 2 36

This equation takes the form of ax 2 + bx + c.

a = 1 b = 0 c = 36

x = [b + / square (b 24ac)] / 2a]

x = [0 + / square (0 24 (1) (36)] / (2) (1)

The discriminator is b 24ac = 144.

i 2 = 1, then Âši 2 = i

Unreal Roots: There are complex roots.

x = [0 + i ÂˆÂš (144)] / (2) (1)

x = [0 i ÂˆÂš (144)] / (2) (1)

x = [0 + i12] / 2

x = [0i12] / 2

6i and 6i are complex roots.

The calculation is as follows:

x 2 (36)

Now you can use the difference of two squares and the square of 36 is 6i:

(x + 6i) (x 6i)

You use two imaginary numbers.

X 2 36

(x + 6) 2 .................... Since 36 is a straight square, just take the square of 6 and square all the terms. It can also be written as (x + 6) (x6). You use real numbers.

x 2 (6i) 2 where i is a dummy number and i 2 = 1

= (x + 6i) (x6i)