Rational, radical, and absolute-value relationships

Recognize the structure. Only the denominator determines the excluded inputs in this rational function.

Work it through. Set the denominator equal to zero:

x² − 9 = (x − 3)(x + 3) = 0.

It vanishes at x = 3 and x = −3. The numerator need not be nonzero for the function to exist.

Answer: Exclude −3 and 3; the domain is R ∖ {−3, 3}.

Check. At x = −1, the numerator is zero but the denominator is −8, so f(−1) = 0 is perfectly valid.

Avoid the trap. A zero numerator gives a zero value when the denominator is nonzero. It does not create a domain restriction.

Recognize the structure. The numerator and denominator share x + 3. Its zero remains excluded after cancellation.

Work it through.

f(x) = $\frac{\text{(x − 3)(x + 3)}}{\text{(x + 3)(x − 2)}}$ = $\frac{\text{x − 3}}{\text{x − 2}}$, x ≠ −3, 2.

A valid zero requires a zero numerator and nonzero denominator. Only x = 3 qualifies.

Answer: (x − 3)/(x − 2) with x ≠ −3, 2; the x-intercept is (3, 0).

Check. The original function is 0/6 = 0 at 3, but is undefined at −3.

Avoid the trap. The canceled numerator zero at −3 creates a hole, not another x-intercept. The original domain controls the graph.

Recognize the structure. The denominators have different factors, so use (x − 1)(x + 1).

Work it through.

$\frac{\text{3(x + 1) + 2(x − 1)}}{\text{(x − 1)(x + 1)}}$ = $\frac{\text{3x + 3 + 2x − 2}}{{\text{x}}^{\text{2}}\text{− 1}}$ = $\frac{\text{5x + 1}}{{\text{x}}^{\text{2}}\text{− 1}}$.

Answer: $\frac{\text{5x + 1}}{{\text{x}}^{\text{2}}\text{− 1}}$, for x ≠ −1, 1.

Check. At x = 0, the original gives −3 + 2 = −1; the combined fraction gives 1/(−1) = −1.

Avoid the trap. Adding the denominators would give 2x, which is not a common-denominator operation and changes the expression.

Recognize the structure. Record x ≠ 2, then multiply by the nonzero denominator.

Work it through.

5 = 3(x − 2) 5 = 3x − 6 x = $\frac{\text{11}}{\text{3}}$.

This candidate is not the excluded value 2.

Answer: 11/3.

Check. The denominator becomes 11/3 − 2 = 5/3, and 5/(5/3) = 3.

Avoid the trap. The denominator is the whole expression x − 2. Multiplying by it requires 3(x − 2), not 3x − 2.

Recognize the structure. The original equation excludes x = 1, and clearing the denominator may produce exactly that value.

Work it through. For x ≠ 1, multiply by x − 1:

x + 2 = 3 x = 1.

But this candidate makes both original denominators zero. It is not a solution.

Answer: 0 real solutions.

Another route. Subtract the right side from the left. On the allowed domain, the difference is (x−1)/(x−1) = 1, so it can never be zero.

Avoid the trap. An equation obtained by clearing denominators is solved only on the original restricted domain. The value 1 cannot be restored by calling the undefined fractions equal.

Recognize the structure. The domain excludes 0; multiplying by x creates a quadratic.

Work it through. On that domain,

x² + 6 = 5x x² − 5x + 6 = 0 (x − 2)(x − 3) = 0.

The candidates are 2 and 3, both allowed.

Answer: 2 and 3.

Check. 2 + 6/2 = 5 and 3 + 6/3 = 5.

Avoid the trap. Multiplying by x affects every term. The right side becomes 5x, and x ⋅ x becomes x².

Recognize the structure. The right side is nonnegative, so squaring can recover the radicand.

Work it through. The original radicand requires x ≥ −4/3. Squaring gives

3x + 4 = 25 3x = 21 x = 7.

The candidate lies in the domain.

Answer: 7.

Check. $\sqrt{\text{3(7) + 4}}$ = $\sqrt{\text{25}}$ = 5.

Avoid the trap. The square root covers the entire expression 3x + 4. Squaring the left side gives 3x + 4, not 3x² + 16.

Recognize the structure. The left side is nonnegative, so any solution must have x ≥ 0.

Work it through. Squaring yields

x + 6 = x² x² − x − 6 = (x − 3)(x + 2) = 0.

The candidates are 3 and −2. The second violates the necessary sign condition x ≥ 0.

Answer: x = 3 only.

Check. At 3, $\sqrt{\text{9}}$ = 3. At −2, $\sqrt{\text{4}}$ = 2 ≠ −2, so the rejected root is genuinely extraneous.

Avoid the trap. The radicand condition x ≥ −6 alone is not enough. The right side must also equal a nonnegative principal square root.

Recognize the structure. The original domain is x ≥ 0. Move one radical so the first squaring has only one on the left.

Work it through. Write $\sqrt{\text{x + 9}}$ = 1 + $\sqrt{\text{x}}$. Squaring gives

x + 9 = 1 + 2$\sqrt{\text{x}}$ + x 8 = 2$\sqrt{\text{x}}$.

Thus $\sqrt{\text{x}}$ = 4, so x = 16.

Answer: 16.

Check. $\sqrt{\text{25}}$ − $\sqrt{\text{16}}$ = 5 − 4 = 1.

Avoid the trap. The square of 1 + $\sqrt{\text{x}}$ is 1 + 2$\sqrt{\text{x}}$ + x. Omitting its cross term destroys the equation. The final check is still required after two squaring steps.

Recognize the structure. The inside expression can be 7 or −7, since both have absolute value 7.

Work it through. Solve the two linear equations:

2x − 5 = 7 x = 6, 2x − 5 = −7 x = −1.

Answer: −1 and 6.

Check. |2(−1) − 5| = | − 7| = 7 and |2(6) − 5| = |7| = 7.

Avoid the trap. Absolute value does not mean “make x positive.” A negative input can produce the required nonnegative output.

Recognize the structure. The right side must satisfy 2x + 1 ≥ 0, so x ≥ −1/2.

Work it through. The two candidate equations are

x − 2 = 2x + 1 x = −3, x − 2 = −(2x + 1) x = $\frac{\text{1}}{\text{3}}$.

The candidate −3 violates x ≥ −1/2; 1/3 survives.

Answer: 1/3.

Check. At x = 1/3, the left side is | − 5/3| = 5/3 and the right side is 5/3. At −3, the right side is negative.

Avoid the trap. The two-branch method creates candidates, not automatic solutions, when the other side contains the variable.

Recognize the structure. The right side is the required distance from 3. Its sign controls the count.

Work it through. If k < 2, the right side is negative, which an absolute value cannot equal. If k = 2, the distance is zero, so x = 3. If k > 2, the positive distance gives two values:

x = 3 + (k − 2) = k + 1, x = 3 − (k − 2) = 5 − k.

They are distinct because k − 2 > 0.

Answer: No solution for k < 2; one for k = 2; two for k > 2.

Check. For k = 5, the distance is 3, giving x = 0 and x = 6, symmetric around 3.

Avoid the trap. At distance zero, the two branches give the same input. Count distinct solutions, not duplicate formulas.

Recognize the structure. Clear the common denominator on the domain x ≠ 2, then check when the resulting root is excluded.

Work it through. Multiply by x − 2:

1 = k + x − 2 x = 3 − k.

This root is forbidden exactly when 3 − k = 2, or k = 1. For every other k, the root is defined and satisfies the equation.

Answer: k = 1.

Check. At k = 1, subtracting 1/(x − 2) from both sides would require 0 = 1, an impossibility on the allowed domain.

Avoid the trap. The absence of a solution comes from the original restriction, not from a failure to solve the resulting linear equation.

Recognize the structure. Collect the terms containing the target b, then inspect the sign of its coefficient.

Work it through. Since a + b > 0, multiply and rearrange:

Ca + Cb = ab Ca = b(a − C).

For a ≠ C, division gives b = Ca/(a − C). Its numerator is positive, so b > 0 requires a − C > 0.

Answer: b = $\frac{\text{Ca}}{\text{a − C}}$, with 0 < C < a.

Check. If a = C > 0, the rearranged equation would require a² = 0, impossible. If C > a, the formula yields a negative b, contrary to the stated condition.

Avoid the trap. A correct algebraic fraction is not the whole answer when the quantities have sign restrictions.

Recognize the structure. The domain excludes x = 2 and x = −1. Clearing both denominators may cancel the apparent quadratic terms.

Work it through. On the allowed domain, multiply by (x − 2)(x + 1):

(x + 1)² + (x − 2)² = 2(x − 2)(x + 1).

Expand both sides: 2x² − 2x + 5 = 2x² − 2x − 4.

Subtracting the identical variable terms leaves 5 = −4, so there is no allowed input.

Answer: 0 real solutions.

Another route. Write A = x + 1 and B = x − 2. The cleared equation is A² + B² = 2AB, or (A − B)² = 0. But A − B = 3, so it would require 9 = 0.

Avoid the trap. Two rational terms do not guarantee a quadratic with two roots. Simplify the entire equation before classifying it.