Example Question 10

Example corresponding to question 10 in college algebra 2.

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?

  1. $x > 7$


  2. $x < 7$


  3. $x < -7$ or $x > 7$


  4. $x \le -7$ or $x \ge 7$


  5. $x=7$

Solution


  1. 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

    $x^2 - 49 \ge 0$

  2. Combining the two conditions above, we can write one single condition as follows

    $x^2 - 49 > 0$

  3. 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