Respuesta :
consider the expression [tex] \frac{x^{2} -x-6}{ x^{2} -4} [/tex]
To factorize the expression in the denominator we use difference of squares: [tex]x^{2} -4=x^{2} - 2^{2} =(x-2)(x+2)[/tex]
To factorize [tex]x^{2} -x-6[/tex] we use the following method:
[tex]x^{2} -x-6=(x-a)(x-b)[/tex]
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 [tex]x^{2} -x-6=(x-a)(x-b)=(x+3)(x-2)[/tex]
[tex] \frac{x^{2} -x-6}{ x^{2} -4}= \frac{(x+3)(x-2)}{(x-2)(x+2)}= \frac{x+3}{x+2} [/tex]
[tex]\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}[/tex]
for x>2
[tex]\frac{1}{x+2}\ \textless \ \frac{1}{2+2}= \frac{1}{4} [/tex]
thus
for x>2,
[tex]1+ \frac{1}{x+2}\ \textless \ 1+ \frac{1}{4}= \frac{5}{4} [/tex]
Answer:
for x>2
[tex]\frac{x^{2} -x-6}{ x^{2} -4} = \frac{x+3}{x+2} \ \textless \ \frac{5}{4} [/tex], (but the expression is never 0)
To factorize the expression in the denominator we use difference of squares: [tex]x^{2} -4=x^{2} - 2^{2} =(x-2)(x+2)[/tex]
To factorize [tex]x^{2} -x-6[/tex] we use the following method:
[tex]x^{2} -x-6=(x-a)(x-b)[/tex]
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 [tex]x^{2} -x-6=(x-a)(x-b)=(x+3)(x-2)[/tex]
[tex] \frac{x^{2} -x-6}{ x^{2} -4}= \frac{(x+3)(x-2)}{(x-2)(x+2)}= \frac{x+3}{x+2} [/tex]
[tex]\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}[/tex]
for x>2
[tex]\frac{1}{x+2}\ \textless \ \frac{1}{2+2}= \frac{1}{4} [/tex]
thus
for x>2,
[tex]1+ \frac{1}{x+2}\ \textless \ 1+ \frac{1}{4}= \frac{5}{4} [/tex]
Answer:
for x>2
[tex]\frac{x^{2} -x-6}{ x^{2} -4} = \frac{x+3}{x+2} \ \textless \ \frac{5}{4} [/tex], (but the expression is never 0)
