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Eddi Din [679]
4 years ago
14

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

Mathematics
1 answer:
Viefleur [7K]4 years ago
8 0
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)
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