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Example 1
What are all the real values of $x$ for which $y$ defined by $y=\dfrac{3}{\sqrt{x^2-49}}$ is a real number?
- $x > 7$
- $x < 7$
- $x < -7$ or $x > 7$
- $x \le -7$ or $x \ge 7$
- $x=7$
Solution
- The expression defining $y$ given above is a ratio of two other algebraic expressions. Division by $0$ in math is not allowed. Hence
$\sqrt{x^2-49} \ne 0$
Also one of the expression includes a square root and the square root of a negative number does not give a real number. Hence
$x^2 - 49 \ge 0$
- Combining the two conditions above, we can write one single condition as follows
$x^2 - 49 > 0$
- We can solve the above inequality by factoring
$(x-7)(x+7) > 0$
$(x-7)(x+7)$ is positive if
A) either $x-7 > 0$ and $x + 7>0$
which give
$x>7$ and $x >-7$
The intersection of $x>7$ and $x >-7$ is $x>7$
B) or $x-7<0$ and $x+7<0$
which give
$x<7$ and $x<-7$
The intersection of $x<7$ and $x<-7$ is $x<-7$
The values of $x$ such that $\dfrac{3}{\sqrt{x^2-49}}$ is a real number are defined as follows:
$x<-7$ or $x>7$
Answer C
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