Question

question 37 of 41
which of the following inequalities represents all values of x for which the quotient below is defined?
\sqrt{24(x - 1)} \div \sqrt{8x^2}
\bigcirc a. x \geq -1
\bigcirc b. x \leq 1
\bigcirc c. x \leq -1
\bigcirc d. x \geq 1

Explanation:

Step1: Analyze the square roots' domains

For a square root \(\sqrt{a}\) to be defined, the radicand \(a\) must be non - negative. Also, for a quotient \(\frac{\sqrt{a}}{\sqrt{b}}\) (or \(\sqrt{a}\div\sqrt{b}\)) to be defined, the denominator \(\sqrt{b}\) must have a non - zero value, so \(b>0\) (since if \(b = 0\), the denominator is zero and the quotient is undefined).

First, consider the numerator \(\sqrt{24(x - 1)}\). The radicand \(24(x - 1)\geq0\). Since \(24>0\), we can divide both sides of the inequality \(24(x - 1)\geq0\) by \(24\) (dividing an inequality by a positive number does not change the direction of the inequality) and get \(x - 1\geq0\), which simplifies to \(x\geq1\) (by adding \(1\) to both sides of the inequality).

Second, consider the denominator \(\sqrt{8x^{2}}\). The radicand \(8x^{2}>0\) (because if \(8x^{2}=0\), then the denominator is zero and the quotient is undefined). Since \(8>0\), we have \(x^{2}>0\). The solution to \(x^{2}>0\) is \(x
eq0\).

Now, we need to find the values of \(x\) that satisfy both conditions. The first condition gives \(x\geq1\) and the second condition gives \(x
eq0\). The values of \(x\) that satisfy \(x\geq1\) also satisfy \(x
eq0\) (because if \(x\geq1\), then \(x\) is at least \(1\), so it can't be \(0\)).

Step2: Check the options

• Option A: \(x\geq - 1\) includes values like \(x = 0\) (where the denominator is zero) and \(x = 0.5\) (where \(x-1=-0.5\) and the numerator's radicand is negative), so this is incorrect.
• Option B: \(x\leq1\) includes values like \(x = 0\) (denominator zero) and \(x = 0.5\) (numerator radicand negative), so this is incorrect.
• Option C: \(x\leq - 1\) has \(x-1\leq - 2\), so the numerator's radicand \(24(x - 1)\leq - 48<0\), so this is incorrect.
• Option D: \(x\geq1\) satisfies both the numerator's radicand non - negativity (\(x-1\geq0\) implies \(24(x - 1)\geq0\)) and the denominator's radicand positivity (\(x^{2}\geq1>0\) when \(x\geq1\), so \(8x^{2}>0\)).

Answer:

D. \(x\geq1\)