When is an expression undefined?

Evaluate each choice.

1. \dfrac{0}{7} = 0, because 0 \times 7 = 0. Zero on top is fine — defined.
2. \dfrac{7}{0}: we would need a number that, times 0, equals 7. Every number times 0 is 0, so no number works — undefined.
3. \sqrt{0} = 0, because 0^2 = 0 — defined.
4. \sqrt[3]{-8} = -2, because (-2)^3 = -8 — defined (cube roots of negatives are real).

Answer: \dfrac{7}{0}

Check: On the TI-30XS, 7 ÷ 0 gives a divide-by-zero error, while the other three give 0, 0 and -2 ✓.

Tip: A zero is only a problem in the denominator. A zero numerator just makes the whole fraction 0.

Evaluate each choice.

1. \sqrt{-16}: we need a number that, times itself, is -16. But 4^2 = 16 and (-4)^2 = 16 — every real number squared is 0 or positive. So \sqrt{-16} is not a real number.
2. -\sqrt{16} = -4: the root is taken first (\sqrt{16} = 4), then the negative is applied. Real.
3. \sqrt[3]{-27} = -3, since (-3)^3 = -27. Real.
4. (-4)^2 = 16. Real.

Answer: \sqrt{-16}

Check: On the TI-30XS, the square root of -16 gives a domain error; the other three give -4, -3 and 16 ✓.

Tip: Look at where the negative sign is. Inside a square root → not real. Outside the root, inside a cube root, or being squared → fine.

1. A fraction is undefined exactly when its denominator equals 0, so simplify every denominator.
2. \dfrac{6}{2 \cdot 3 - 6}: 2 \cdot 3 - 6 = 6 - 6 = 0, so this is \dfrac{6}{0} — undefined.
3. \dfrac{2 \cdot 3 - 6}{6}: the numerator is 0, so the value is \dfrac{0}{6} = 0 — defined.
4. \dfrac{6}{3 - 6} = \dfrac{6}{-3} = -2 — defined.
5. \dfrac{-6}{-6} = 1 — defined.

Answer: \dfrac{6}{2 \cdot 3 - 6}

Check: Ask "what times 0 gives 6?" No number does, so \tfrac{6}{0} has no value ✓. The other three each have a real value: 0, -2, 1.

Tip: Always simplify the denominator before deciding. A denominator can look harmless (2 \cdot 3 - 6) and still equal 0. A zero in the numerator is never a problem.

1. An even root (square root, 4th root, …) of a negative number is not real, because any real number raised to an even power is zero or positive.
2. \sqrt[4]{-16}: we would need a number whose 4th power is -16. But 2^4 = 16 and (-2)^4 = 16 — never -16. Not real.
3. \sqrt[3]{-125} = -5, because (-5)^3 = -125. Real (odd root).
4. -\sqrt[4]{16} = -2. The negative sign is outside the root, so we take \sqrt[4]{16} = 2 and then make it negative. Real.
5. \sqrt{(-4)^2} = \sqrt{16} = 4. Square first: (-4)^2 = 16. Real.

Answer: \sqrt[4]{-16}

Check: On the calculator, (-2)^4 = 16 and 2^4 = 16, so no real number has 4th power -16 ✓.

Tip: Ask two questions: Is the root even? and Is the number inside negative? Only "yes" to both makes the expression not real.

1. Set only the denominator equal to 0: x + 7 = 0.
2. Subtract 7 from both sides: x = -7.
3. The numerator x - 2 does not matter. At x = 2 the expression is \dfrac{0}{9} = 0, which is defined.

Answer: x = -7 only

Check: At x = -7: \dfrac{-7 - 2}{-7 + 7} = \dfrac{-9}{0} → undefined ✓.

Tip: Substitute your answer into the denominator to catch sign slips. x + 7 = 0 gives x = -7, not 7 (since 7 + 7 = 14).

1. \sqrt{x - 6} is not real when the radicand is negative: x - 6 < 0.
2. Add 6 to both sides: x < 6.
3. Note: at x = 6, \sqrt{6 - 6} = \sqrt{0} = 0, which is real. So 6 itself is allowed.

Answer: all x < 6

Check: Try x = 2: \sqrt{2 - 6} = \sqrt{-4} — not real ✓. Try x = 10: \sqrt{4} = 2 — real ✓. Try x = 6: \sqrt{0} = 0 — real ✓.

Tip: Test one number from each side of the boundary. Picking an easy value like x = 0 tells you immediately which side is the "bad" side.

1. The formula is undefined when the denominator is 0: n - 3 = 0.
2. Solve: n = 3.
3. Meaning: if only 3 people come, they are all hosts, so nobody is paying — there is no "cost per paying guest".

Answer: n = 3

Check: C = \dfrac{360}{3 - 3} = \dfrac{360}{0} → undefined ✓. With n = 12, C = \tfrac{360}{9} = 40, a sensible $40 each.

Tip: In a real-world formula, the "undefined" value usually describes a situation that cannot happen or makes no sense (no paying guests, zero time, zero people). Setting the denominator to 0 finds it.

1. Simplify inside the cube root: 2 - 29 = -27.
2. \sqrt[3]{-27} = -3, because (-3)^3 = (-3)(-3)(-3) = -27. Cube roots of negatives are real.
3. Simplify inside the square root: 25 - 16 = 9.
4. \sqrt{9} = 3.
5. Add: -3 + 3 = 0.

Answer: 0

Check: (-3)^3 = -27 ✓ and 3^2 = 9 ✓, so -3 + 3 = 0 ✓.

Tip: A negative number inside a cube root is fine — the answer is just negative. Also, never split a root across subtraction: \sqrt{25 - 16} is \sqrt{9} = 3, not 5 - 4 = 1.

1. \sqrt{3x - 12} is not real exactly when the radicand is negative: 3x - 12 < 0.
2. Add 12 to both sides: 3x < 12.
3. Divide both sides by 3 (a positive number, so the sign stays the same): x < 4.

Answer: x < 4

Check: Test x = 0: \sqrt{3(0) - 12} = \sqrt{-12} — not real ✓. Test x = 4: \sqrt{0} = 0 — real, so 4 is not included ✓. Test x = 5: \sqrt{3} \approx 1.73 — real ✓.

Tip: The boundary value (here x = 4) makes the radicand 0, and \sqrt{0} = 0 is real. So "not real" uses a strict <, never \le. Test one number on each side to confirm the direction.

1. The formula is a fraction, so it is undefined when the denominator is 0: t_2 - t_1 = 0.
2. That happens when t_2 = t_1 — the two times are equal.
3. Check each choice:
   • d = 0, t_1 = 1.5, t_2 = 4: \dfrac{0}{2.5} = 0 → defined.
   • d = 2, t_1 = 0, t_2 = 4: \dfrac{2}{4} = 0.5 → defined.
   • d = 2, t_1 = 2.5, t_2 = 2.5: \dfrac{2}{0} → undefined.
   • d = 2, t_1 = 4, t_2 = 0: \dfrac{2}{-4} = -0.5 → defined (a strange reading, but a real number).

Answer: d = 2, t_1 = 2.5, t_2 = 2.5

Check: 2.5 - 2.5 = 0, so the speed would be \tfrac{2}{0} — no number works ✓. In real life, a cart cannot travel 2 meters in 0 seconds.

Tip: In a formula, the variable that makes the formula undefined is whatever makes the whole denominator zero. A zero time reading (t_1 = 0) is not the same as a zero difference.

1. Slope = \dfrac{\text{rise}}{\text{run}} = \dfrac{y_2 - y_1}{x_2 - x_1}. It is undefined when the run is 0, which means every point has the same x-coordinate — a vertical line.
2. Line B passes through (-3, -6) and (-3, 6): slope = \dfrac{6 - (-6)}{-3 - (-3)} = \dfrac{12}{0} → undefined.
3. Line A passes through (-6, 4) and (6, 4): slope = \dfrac{0}{12} = 0 (horizontal).
4. Line C passes through (0, 0) and (5, 5): slope = \dfrac{5}{5} = 1.
5. Line D passes through (0, 1) and (3, -5): slope = \dfrac{-6}{3} = -2.

Answer: Line B

Check: Pick any two points on line B; their x-values are both -3, so the run is always 0 ✓.

Tip: Memory trick — a horizontal line is flat ground: slope 0. A vertical line is a wall you cannot walk up: slope undefined.

1. Set the denominator equal to 0: x^2 + 2x - 15 = 0.
2. Factor: find two numbers that multiply to -15 and add to +2. They are 5 and -3. So x^2 + 2x - 15 = (x + 5)(x - 3).
3. Set each factor to 0: x + 5 = 0 \Rightarrow x = -5; x - 3 = 0 \Rightarrow x = 3.

Answer: x = -5 or x = 3

Check: (-5)^2 + 2(-5) - 15 = 25 - 10 - 15 = 0 ✓. 3^2 + 2(3) - 15 = 9 + 6 - 15 = 0 ✓.

Tip: The numerator x + 3 does not matter — at x = -3 the expression is \tfrac{0}{-18} = 0, which is defined. After factoring, each factor (x + a) gives x = -a: the sign flips.

1. The formula is undefined when its denominator is 0. The denominator is g, the gallons added.
2. So the formula breaks down when g = 0 — Bea added no gas.
3. The other situations:
   • m_2 = m_1: the numerator is 0, so mpg = \tfrac{0}{g} = 0 (if g is not 0). Defined.
   • m_1 = 0 (a brand-new car): the numerator is just m_2. Defined.
   • m_2 is larger than m_1 by 300: mpg = \tfrac{300}{g}. Defined.

Answer: She added 0 gallons (g = 0).

Check: With m_2 - m_1 = 300 and g = 0: \tfrac{300}{0} — no number times 0 gives 300 ✓. It also makes sense: "miles per gallon" means nothing if no gallons were used to refill.

Tip: In a word problem, find the variable in the denominator first. Equal readings or zero readings in the numerator give 0 or an ordinary number, not "undefined".

Test each statement.

1. "\dfrac{0}{9} is undefined." False — \dfrac{0}{9} = 0, since 0 \times 9 = 0.
2. "\sqrt[3]{-125} is not a real number." False — (-5)^3 = -125, so \sqrt[3]{-125} = -5.
3. "\sqrt{x + 2} is a real number when x = -2." True — \sqrt{-2 + 2} = \sqrt{0} = 0, which is real.
4. "\dfrac{x}{x + 2} is undefined when x = 0." False — \dfrac{0}{2} = 0. It is undefined at x = -2.

Answer: \sqrt{x + 2} is a real number when x = -2.

Check: The only things that are undefined are a 0 denominator or a negative under an even root. Statement 3 has neither: the radicand is exactly 0 ✓.

Tip: Zero is only dangerous in one place: the denominator. Zero under a square root, or on top of a fraction, gives the ordinary answer 0.

The expression is real only if both conditions hold: x + 5 \ge 0 (root) and x - 1 \ne 0 (denominator).

1. x = -10: radicand -10 + 5 = -5 → \sqrt{-5} not real ✗.
2. x = -6: radicand -6 + 5 = -1 → \sqrt{-1} not real ✗.
3. x = -5: radicand 0, so \sqrt{0} = 0; denominator -5 - 1 = -6. Value \dfrac{0}{-6} = 0 → real ✓.
4. x = 1: radicand 6 is fine, but denominator 1 - 1 = 0 → undefined ✗.

Answer: x = -5

Check: \dfrac{\sqrt{-5 + 5}}{-5 - 1} = \dfrac{0}{-6} = 0, a real number ✓.

Tip: When an expression has both a root and a fraction, check both rules for each value. And remember that a zero under the root, or a zero on top of the fraction, is perfectly fine.

The expression needs two things: x - 2 \ge 0 (the root is real) and \sqrt{x - 2} \ne 0 (the denominator is not 0). Together: x - 2 > 0, so x > 2.

Test each choice:

1. x = 2: denominator \sqrt{2 - 2} = \sqrt{0} = 0, so we get \dfrac{-2}{0} → undefined.
2. x = 3: \dfrac{3 - 4}{\sqrt{1}} = \dfrac{-1}{1} = -1 → defined.
3. x = 4: \dfrac{0}{\sqrt{2}} = 0 → defined.
4. x = 6: \dfrac{2}{\sqrt{4}} = \dfrac{2}{2} = 1 → defined.

Answer: x = 2

Check: \sqrt{0} = 0 is a real number, but putting it in the denominator gives division by zero ✓.

Tip: \sqrt{0} is usually harmless — but not when it is the whole denominator. When a root sits in a denominator, the radicand must be strictly greater than 0.

1. Step 1 is correct: undefined means the denominator is 0.
2. Step 2 is correct: x^2 - 36 + 36 = 0 + 36 gives x^2 = 36.
3. Step 3 is the first error: x^2 = 36 has two solutions, x = 6 and x = -6, because (-6)^2 = 36 too.
4. Step 4 is wrong only because it repeats the mistake from Step 3. The correct conclusion: the expression is undefined at x = 6 and x = -6.

Answer: Step 3

Check: (-6)^2 - 36 = 36 - 36 = 0, so x = -6 also makes the denominator 0 ✓.

Tip: In "find the error" questions, test each step by itself and stop at the first wrong one. With x^2 = k (positive k), always write x = \pm\sqrt{k}.

1. The formula is undefined when the denominator is 0: 12 - r = 0, so r = 12.
2. Meaning: when r = 12, water drains out at 12 gallons per minute while the hose adds 12 gallons per minute. The net fill rate is 12 - 12 = 0 gallons per minute.
3. At a net rate of 0, the water level never rises, so the pool never fills. No finite time works, which matches \tfrac{V}{0} having no value.
4. (For comparison: r = 0 means no leak, and T = \tfrac{V}{12}, an ordinary time.)

Answer: r = 12; the leak drains as fast as the hose fills, so the pool never fills.

Check: With V = 600 and r = 10: T = \tfrac{600}{2} = 300 min. With r = 11.9: T = \tfrac{600}{0.1} = 6000 min — the time grows huge as r approaches 12 ✓.

Tip: On the GED, an "undefined" value in a real formula usually means a situation that cannot happen or never ends. Explain it with the units of the denominator.