Question

determine the solution for ( x^2 + 36 > 12x )
( circ ) ( {x | x = 6} )
( circ ) ( {x | -6 < x < 6} )
( circ ) ( {x | x in mathbb{r} } )
( circ ) ( {x | x in mathbb{r} \text{ and } x
eq 6} )
( circ ) ( emptyset )
the solution for ( x^2 + 2x + 8 leq 0 ) is
done

Explanation:

Step1: Rearrange the inequality

Start with \( x^{2}+36>12x \). Subtract \( 12x \) from both sides to get \( x^{2}-12x + 36>0 \).

Step2: Factor the quadratic

Notice that \( x^{2}-12x + 36=(x - 6)^{2} \). So the inequality becomes \( (x - 6)^{2}>0 \).

Step3: Analyze the square

A square of a real number \( (x - 6)^{2} \) is always non - negative. It is equal to 0 when \( x = 6 \) and positive otherwise. So \( (x - 6)^{2}>0 \) when \( x
eq6 \) and \( x\in\mathbb{R} \).

For the second inequality \( x^{2}+2x + 8\leq0 \):

Step1: Complete the square or use discriminant

First, find the discriminant of the quadratic \( ax^{2}+bx + c \) (here \( a = 1 \), \( b = 2 \), \( c = 8 \)) using \( D=b^{2}-4ac \).
\( D=(2)^{2}-4\times1\times8=4 - 32=- 28<0 \).
Since \( a = 1>0 \), the quadratic function \( y=x^{2}+2x + 8 \) opens upwards and is always positive (because the discriminant is negative, it has no real roots and is always above the x - axis). So the inequality \( x^{2}+2x + 8\leq0 \) has no solution.

Answer:

For \( x^{2}+36>12x \): \(\{x|x\in\mathbb{R}\text{ and }x
eq6\}\)
For \( x^{2}+2x + 8\leq0 \): No solution