Unknown-constant and solution-count problems

Recognize. An equation has infinitely many solutions when it simplifies to a true statement independent of x.

Work. Subtract 3x + 7 from both sides to obtain (k − 3)x = 0. For every real x to satisfy this, require k − 3 = 0, giving k = 3.

Answer: 3.

Check. Substitution gives 3x + 7 = 3x + 7, true for all real x. If k ≠ 3, the only solution would be x = 0.

Avoid the trap. A coefficient becoming zero is not enough by itself in every problem. Here the constants cancel too, which is what creates an identity.

Recognize. Compare this equation with the preceding one: the variable coefficients can match, but the constant terms are different.

Work. Rearranging gives (k − 3)x = −5. At k = 3 this becomes 0 = −5, an impossible statement. For every k ≠ 3, division gives one solution, x = −5/(k − 3).

Answer: 3.

Check. With k = 3, subtracting 3x from the original equation would require 7 = 2.

Avoid the trap. “The x terms disappear” does not tell you whether the result is an identity or a contradiction. Read the remaining constants.

Recognize. The second equation’s x coefficient is three times the first. Match the y coefficient to make the lines parallel, then compare constants.

Work. Multiplying the first equation by 3 gives 6x + 15y = 33. Thus k = 15 makes the second equation 6x + 15y = 30, which contradicts the first.

Answer: 15.

Check. The same left side cannot equal both 33 and 30 at one point. If k ≠ 15, subtracting three times the first equation leaves (k − 15)y = −3, which determines one y and then one x.

Avoid the trap. Matching coefficient ratios creates infinitely many solutions only when the constant ratio matches as well.

Recognize. The equality is an identity, so matching coefficients is valid.

Work. Expand the left side as ax² + (b − 2a)x − 2b. The quadratic coefficient gives a = 4, and the constant term gives −2b = 6, so b = −3. Their sum is 1.

Answer: 1.

Check. The remaining coefficient is b − 2a = −3 − 8 = −11, matching the given linear term.

Alternative. Setting x = 1 gives −(a + b) = 4 − 11 + 6 = −1, so the target is 1 directly.

Avoid the trap. The constant term is −2b, not b. An identity requires every coefficient, including signs, to be consistent.

Recognize. A known root can be substituted into the original polynomial. The result is an equation in the parameter only.

Work. Substitute 3: 9 + 3k − 24 = 0. Thus 3k = 15, so k = 5.

Answer: 5.

Check. At k = 5, the polynomial factors as (x − 3)(x + 8), confirming that 3 is a root.

Avoid the trap. The fact that 3 is a solution does not mean it is the only solution. The problem uses one root to determine the constant, not to specify the entire solution set.

Recognize. The leading coefficient is 1, so the equation is quadratic for every k. Exactly one real root means the discriminant is zero.

Work. Set (−8)² − 4(1)k = 0, giving 64 − 4k = 0 and k = 16.

Answer: 16.

Check. The equation becomes x² − 8x + 16 = (x − 4)² = 0, whose only distinct real solution is 4.

Alternative. Complete the square first: (x − 4)² = 16 − k. Exactly one real solution occurs when the right side is zero.

Avoid the trap. A repeated factor represents one distinct solution, not two different real inputs.

Recognize. Two distinct real solutions require a positive, not merely nonnegative, discriminant.

Work. Require 6² − 4k > 0, so 36 − 4k > 0 and k < 9. The greatest integer satisfying this is 8.

Answer: 8.

Check. At k = 8, the polynomial is (x + 2)(x + 4), giving two distinct roots. At k = 9, it is (x + 3)², giving only one distinct root.

Avoid the trap. Using Δ ≥ 0 would include the repeated-root boundary and give an incorrect greatest integer.

Recognize. Equate the two outputs. Exactly one intersection occurs when the resulting quadratic has one real root.

Work. Setting x² + 2x + 3 = 4x + k gives x² − 2x + (3 − k) = 0. Its discriminant is 4 − 4(3 − k) = 4k − 8. Set this equal to zero to obtain k = 2.

Answer: 2.

Check. At k = 2, the equation is (x − 1)² = 0. The common point is (1, 6), and there is only one intersection input.

Avoid the trap. The parameter k is the line’s intercept, not the intersection’s y coordinate. Here those values are 2 and 6, respectively.

Recognize. The circle’s center is (2, −1) and its radius is 5. Horizontal tangents lie at the highest and lowest points.

Work. Their y coordinates are −1 + 5 = 4 and −1 − 5 = −6. Therefore the product of the two k values is 4(−6) = −24.

Answer: −24.

Check. Substituting either y = 4 or y = −6 leaves (x − 2)² = 0, giving exactly one point on that horizontal line.

Avoid the trap. The radius is 5, not 25, and the center’s y coordinate is −1, not 1. Read both features before calculating.

Recognize. Dividing the observations removes a and leaves the growth over two input units.

Work. From ab² = 18 and ab⁴ = 162, division gives b² = 9. Since b > 0, b = 3. Then a(9) = 18, so a = 2 and a + b = 5.

Answer: 5.

Check. The model f(t) = 2 · 3^t gives 18 at t = 2 and 162 at t = 4.

Avoid the trap. The ratio 9 is the multiplier over two units, not the one-unit base. The positivity condition rules out the negative square root when recovering b.

Recognize. The leading coefficient can vanish at k = 1, so a discriminant-only solution may miss a case.

Work. If k = 1, the equation becomes 2x + 1 = 0, with one solution. If k ≠ 1, it is quadratic; require 2² − 4(k − 1)(1) = 0. This gives 8 − 4k = 0, so k = 2.

Answer: k = 1 or k = 2.

Check. At k = 1, the root is −1/2. At k = 2, the equation is (x + 1)² = 0, with the single distinct root −1.

Avoid the trap. Dividing by k − 1 at the start would discard the linear case. The phrase “exactly one real solution” does not guarantee the equation remains quadratic.

Recognize. The original equation excludes x = 3. After simplifying, check whether the candidate lands on that forbidden input.

Work. For x ≠ 3, the equation becomes x + 3 = k, so its only candidate is x = k − 3. This is forbidden exactly when k − 3 = 3, or k = 6.

Answer: 6.

Check. With k = 6, the simplified equation requires x = 3, where the original denominator is zero. Every other real k gives one allowed input.

Avoid the trap. The simplified line has a missing input inherited from the rational expression. Ignoring the exclusion incorrectly predicts one solution for every k.

Recognize. Squaring and checking roots separately for every parameter is cumbersome. Set u = $\sqrt{\text{x + 4}}$, so u ≥ 0 and x = u² − 4. Each nonnegative u corresponds to exactly one allowed x.

Work. The original equation becomes u = u² − 4 − k, or

u² − u − (k + 4) = 0.

Its discriminant is 4k + 17, its root sum is 1, and its root product is −(k + 4). Count only nonnegative roots:

Answer: k = −17/4 or k > −4.

Check. At k = −17/4, u = 1/2 gives x = −15/4, and both sides of the original equation equal 1/2. At k = −4, the solutions x = −4, −3 are both valid, so this boundary is excluded from the one-solution set.

Avoid the trap. A positive discriminant counts two real roots of the transformed quadratic, not necessarily two valid roots of the original radical equation. The condition u ≥ 0 is essential.

Transfer. Substitution can make a domain restriction easier to count. The target is not a memorized radical rule but the intersection of algebraic roots with the permitted domain.

Recognize. First find the intersection inputs as expressions in k. Then apply the strict condition x > 0; the system may still have another intersection elsewhere.

Work. Equate outputs: x² = 2kx − k² + 9 (x − k)² = 9.

The two distinct inputs are k − 3 and k + 3. Exactly one is positive when the larger is positive and the smaller is not: k + 3 > 0 and k − 3 ≤ 0.

Thus −3 < k ≤ 3. The integer values are −2, −1, 0, 1, 2, 3, a total of six.

Answer: 6.

Check. At k = −3, the inputs are −6 and 0, so neither is positive. At k = 3, the inputs are 0 and 6, so exactly one is positive. For k > 3, both inputs are positive.

Avoid the trap. “Exactly one intersection point has positive x” does not mean “the graphs have exactly one intersection.” In fact, these graphs always have two distinct real intersections.

Transfer. Solve the geometric system first, then filter. Keep a boundary at zero separate from the positive region; changing > to ≥ changes the admissible parameter endpoints.

A language distinction worth protecting

Count the objects named by the question. “One real root,” “one positive root,” and “one intersection in the first quadrant” impose different conditions, even on the same equations.

Recognize. Three properties must hold together: two distinct real roots, a positive sum, and a positive product. The leading coefficient is always 1, so no lower-degree case is needed.

Work. The discriminant condition is

(−2k)² − 4(k + 2) = 4(k² − k − 2) = 4(k − 2)(k + 1) > 0,

which gives k < −1 or k > 2. The root sum is 2k, so it must be positive: k > 0. The product is k + 2, requiring k > −2. Intersecting all three conditions leaves k > 2. Within the stated integer range, the possibilities are k = 3, 4, 5, 6.

Answer: 4.

Check. At k = 2, the equation is (x − 2)² = 0, so the roots are not distinct. At k = 3, the roots are 1 and 5, both positive and distinct. Negative values that make the discriminant positive fail the positive-sum requirement.

Avoid the trap. A positive discriminant alone says nothing about both roots being positive. A positive product alone could mean two negative roots. The requested property is the intersection, not the union, of the conditions.

Transfer. When several adjectives modify “solutions,” translate each adjective separately. Then combine the conditions only after their meanings are explicit.