Question

Determine the solution for x2 + 36 > 12x

Answer 1

For this case we have the following inequality:

`x ^ 2 + 36> 12x`

We subtract 12x on both sides of the inequality:

`x ^ 2-12x + 36> 0`

We convert to an equation:

`x ^ 2-12x + 36 = 0`

To factor the equation, we must find two numbers that when multiplied give +36 and when added together obtain -12. These numbers are -6 and -6.

`-6-6 = -12\\-6 * -6 = + 36`

So, we have:

`(x-6) (x-6) = 0`

The solution is
`x = 6`

We create test intervals:

`x <6`

We test a value in the interval, such as x = 0:

`0 ^ 2 + 36> 12 (0)\\36> 0`

Is fulfilled!

`x> 6`

We test a value in the interval, such as x = 10.

`10 ^ 2 + 36> 12 (10)\\100 + 36> 120\\136> 120`

Is fulfilled!

Thus, the solution is given by:

`x <6`or
`x> 6`

Answer:

`x <6`or
`x> 6`

Answer 2

solution for x^2 + 36 > 12x is x > 6 or x< 6

Explanation:

We need to solve the equation:

x^2 + 36 > 12x

Subtract 12x from both sides

x^2 + 36 - 12x > 0

Rearranging

x^2 -12x + 36 > 0

Now factorizing:

x^2 -6x -6x + 36 >0

x(x-6)-6(x-6) > 0

(x-6)(x-6) > 0

(x-6)^2 > 0

For, x^n > 0 and n is even then x >0 or x<0

so,

x-6 >0 or x-6 <0

x > 6 or x< 6

So, solution for x^2 + 36 > 12x is x > 6 or x< 6