Question

If x>2, then x^2-x-6/x^2-4=

Answer 1

Consider the expression $\frac{x^{2} -x-6}{ x^{2} -4}$

To factorize the expression in the denominator we use difference of squares: $x^{2} -4=x^{2} - 2^{2} =(x-2)(x+2)$

To factorize $x^{2} -x-6$ we use the following method:

$x^{2} -x-6=(x-a)(x-b)$

where a, b are 2 numbers such that a+b= -1, the coefficient of x,

and a*b= -6, the constant.

such 2 numbers can be easily checked to be -3 and 2

(-3*2=6, -3+2=-1)

So $x^{2} -x-6=(x-a)(x-b)=(x+3)(x-2)$

$\frac{x^{2} -x-6}{ x^{2} -4}= \frac{(x+3)(x-2)}{(x-2)(x+2)}= \frac{x+3}{x+2}$


$\frac{x+3}{x+2}= \frac{x+2+1}{x+2}= \frac{x+2}{x+2}+ \frac{1}{x+2}=1+ \frac{1}{x+2}$

for x>2

$\frac{1}{x+2}\ \textless \ \frac{1}{2+2}= \frac{1}{4}$

thus

for x>2, 

$1+ \frac{1}{x+2}\ \textless \ 1+ \frac{1}{4}= \frac{5}{4}$


Answer: 

for x>2

$\frac{x^{2} -x-6}{ x^{2} -4} = \frac{x+3}{x+2} \ \textless \ \frac{5}{4}$, (but the expression is never 0)