Quadratic equations and functions

Recognize the structure. The numbers −3 and −4 have product 12 and sum −7.

Work it through. Factor the expression and use the zero-product property:

(x − 3)(x − 4) = 0 x = 3 or x = 4.

Answer: 3 and 4.

Check. Substituting gives 9 − 21 + 12 = 0 and 16 − 28 + 12 = 0.

Avoid the trap. The constants in the factors are −3 and −4, but the roots are the inputs that make those factors zero: positive 3 and 4.

Recognize the structure. The equation is already centered around a squared expression.

Work it through. Divide by 3, then account for both square roots:

(x − 2)² = 16 x − 2 = ±4.

The two solutions are x = 6 and x = −2.

Answer: The larger solution is 6.

Check. 3(6 − 2)² = 3 ⋅ 16 = 48. The other solution also satisfies the original equation, but is not the requested one.

Avoid the trap. Writing only x − 2 = 4 would happen to produce the larger root here, but it is not a complete solution process and would fail a question asking for both roots.

Recognize the structure. The leading coefficient is 2, so a factor can create a fractional root.

Work it through. Factor by splitting the middle term:

2x² + 6x − x − 3 = (2x − 1)(x + 3) = 0.

Thus 2x − 1 = 0 or x + 3 = 0, giving x = 1/2 or x = −3.

Answer: 1/2.

Check. At x = 1/2, the original is 1/2 + 5/2 − 3 = 0.

Avoid the trap. A factor 2x − 1 has zero 1/2, not 1 or 2. Solve the linear factor rather than reading off its constant alone.

Recognize the structure. Half of the linear coefficient is 3, so adding 9 creates a square.

Work it through. x² + 6x = −2 x² + 6x + 9 = 7.

Therefore (x + 3)² = 7, giving x + 3 = ±$\sqrt{\text{7}}$.

Answer: x = −3 − $\sqrt{\text{7}}$ and x = −3 + $\sqrt{\text{7}}$.

Check. Substituting either root into (x + 3)² = 7 makes the equation true. Every preceding step was reversible.

Avoid the trap. Add the same 9 to both sides. Adding it only on the left changes the equation and its roots.

Recognize the structure. The equation does not factor conveniently over the integers. Use a = 3, b = −4, and c = −2.

Work it through.

x = $\frac{\text{4 ±}$ \sqrt{\text{(−}{\text{4)}}^{\text{2}}\text{− 4(3)(−2)}}$}{\text{6}}$ = $\frac{\text{4 ±}$ \sqrt{\text{40}}$}{\text{6}}$ = $\frac{\text{2 ±}$ \sqrt{\text{10}}$}{\text{3}}$.

The discriminant is 40 > 0, so there are two distinct real roots.

Answer: $\frac{\text{2 −}$ \sqrt{\text{10}}$}{\text{3}}$ and $\frac{\text{2 +}$ \sqrt{\text{10}}$}{\text{3}}$.

Check. The root sum is 4/3 = −b/a, and the product is (4 − 10)/9 = −2/3 = c/a.

Avoid the trap. Since b = −4, the numerator begins with −b = 4. The radical and the 4 are both divided by 6.

Recognize the structure. The negative coefficient makes the squared term nonpositive.

Work it through. For every real x, (x + 1)² ≥ 0, so −2(x + 1)² ≤ 0. The largest possible output is therefore 18, achieved when the square is zero: x + 1 = 0 x = −1.

Answer: Maximum value 18, at x = −1.

Check. f(−1) = 18. Any other input makes the square positive and subtracts a positive quantity from 18.

Avoid the trap. The +1 inside the parentheses corresponds to a vertex input of −1, not 1. Also, the maximum value is not the input where it occurs.

Recognize the structure. Factor out 2 from the quadratic and linear terms before completing the square.

Work it through. f(x) = 2(x² − 6x) + 11 = 2((x − 3)² − 9) + 11 = 2(x − 3)² − 7.

The squared term is nonnegative, so the minimum is −7, attained at x = 3.

Answer: −7.

Check. f(3) = 18 − 36 + 11 = −7.

Avoid the trap. Adding 9 inside a factor of 2 changes the total by 18, not 9. The outside coefficient must be accounted for.

Recognize the structure. The roots determine two factors; the extra value determines the vertical scale.

Work it through. Write f(x) = a(x + 2)(x − 6). Substituting x = 0 gives −12a = −12, so a = 1. The axis of symmetry is halfway between the roots:

x = $\frac{\text{−2 + 6}}{\text{2}}$ = 2.

Since a > 0, the vertex gives a minimum. Its output is f(2) = (4)(−4) = −16.

Answer: −16.

Check. Expanding gives x² − 4x − 12 = (x − 2)² − 16, which displays the same minimum.

Avoid the trap. Knowing the zeros alone does not determine the scale a. Use the additional point before evaluating the vertex.

Recognize the structure. Use the vertex to choose f(x) = a(x − 3)² − 4.

Work it through. The point (1, 4) gives 4 = 4a − 4, so a = 2. Then f(0) = 2(0 − 3)² − 4 = 18 − 4 = 14.

Answer: 14.

Check. The derived rule reproduces both labeled points. The intercept lies above the visible graph window, so the picture alone cannot be read for the answer.

Avoid the trap. The visible top of a graph is not a maximum; the axes’ window is only part of the plane.

Recognize the structure. Ground level means height zero, and time after release must satisfy t ≥ 0.

Work it through. Set h(t) = 0 and divide by −5:

t² − 4t − 5 = (t − 5)(t + 1) = 0.

The algebraic roots are 5 and −1. Only 5 is in the stated time domain.

Answer: 5 seconds.

Check. h(5) = −125 + 100 + 25 = 0. The negative root refers to an extrapolated time before release, not the requested event.

Avoid the trap. A zero of the height function is a time, so the answer’s units are seconds, not meters.

Recognize the structure. The quadratic coefficient is 1 ≠ 0, so exactly one real root means discriminant zero.

Work it through. D = (−8)² − 4(1)(k) = 64 − 4k.

Set D = 0: 64 − 4k = 0, so k = 16.

Answer: 16.

Another route. Complete the square: x² −8x +k = (x −4)² +k −16. Exactly one root occurs when the square must equal 0, which again requires k = 16.

Check. The equation becomes (x − 4)² = 0, whose only distinct solution is 4.

Avoid the trap. D > 0 creates two distinct roots; D ≥ 0 is too broad for a question asking for exactly one.

Recognize the structure. Use a strict discriminant inequality, then apply the integer requirement.

Work it through. No real roots requires

k² − 36 < 0 k² < 36 −6 < k < 6.

The greatest integer in this open interval is 5.

Answer: 5.

Check. For k = 5, the discriminant is 25 − 36 = −11 < 0. At k = 6, it is 0, giving a real repeated root and violating the requirement.

Avoid the trap. The endpoints are excluded. Rounding the boundary 6 does not produce an allowed parameter.

Recognize the structure. The target is (r + s)² − 2rs, so the individual roots are unnecessary.

Work it through. From the coefficients, r + s = 11/3 and rs = 4/3. Therefore

r² + s² = ($\frac{\text{11}}{\text{3}}$)² − 2 ($\frac{\text{4}}{\text{3}}$) = $\frac{\text{121}}{\text{9}}$ − $\frac{\text{24}}{\text{9}}$ = $\frac{\text{97}}{\text{9}}$.

Answer: 97/9.

Check. The discriminant is 73 > 0, so the equation indeed has two distinct real roots. The sum-product identities follow by expanding 3(x − r)(x − s).

Avoid the trap. The sum is −b/a = 11/3, not −11/3; the product is c/a = 4/3, not 4.

Recognize the structure. Revenue is price times quantity, and the input is restricted to whole numbers.

Work it through. The model is

R(n) = (20 + 2n)(120 − 8n) = 2400 + 80n − 16n².

Its continuous vertex occurs at n = −80/[2(−16)] = 2.5. Compare the adjacent feasible integers:

R(2) = 24 ⋅ 104 = 2496, R(3) = 26 ⋅ 96 = 2496.

Because the parabola opens downward, all other feasible integers are farther from the vertex and yield less revenue.

Answer: $2,496.

Avoid the trap. The continuous maximum is $2,500 at n = 2.5, but half of a permitted price-increase step is not an allowed choice.

Recognize the structure. The leading coefficient can vanish, so split off that case before using the discriminant.

Work it through. Case 1: k = 1. The equation becomes 4x + 4 = 0, which has the one solution x = −1. Thus k = 1 qualifies.

Case 2: k ≠ 1. The equation is quadratic. Exactly one real solution requires

D = 4² − 4(k − 1)(4) = 32 − 16k = 0,

so k = 2. This value is allowed in Case 2. The qualifying parameters are 1 and 2.

Answer: k = 1 or k = 2; their sum is 3.

Check. At k = 2, the equation is (x+2)² = 0. At k = 1, it is linear with one solution. Both satisfy the requested solution count, for different reasons.

Avoid the trap. Using only D = 0 would find k = 2 and miss the valid linear case k = 1. A formula’s hypotheses must be checked before it is applied.