Algebra Quiz: Quadratic Equations

Factoring 3x²−13x−10 requires numbers multiplying to −30 and summing to −13. The numbers −15 and 2 satisfy this condition. Rewriting gives (3x+2)(x−5)=0. Setting each factor equal to zero produces x=−2/3 and x=5. Both satisfy the original equation when substituted. This confirms the factorization and verifies the solution values algebraically without approximation.

Break-even occurs when profit equals zero. Setting −2x²+13x−15=0 and factoring gives (2x−3)(x−5)=0. Solutions are x=1.5 and x=5. Since x represents thousands of units, multiply by 1000. The company must sell 1500 or 5000 items to break even. Both points indicate zero profit and define realistic production thresholds.

The ball hits the ground when height equals zero. Setting −5t²+35t=0 and factoring yields t(−5t+35)=0. Solutions are t=0 and t=7. Time zero represents launch, so the physical solution is t=7 seconds. This result reflects the symmetry of quadratic motion and confirms when the height returns to ground level after ascent.

To find impact time, set height to zero: −5t²+80=0. Rearranging gives t²=16. Taking square roots yields t=±4. Negative time is not physically meaningful, so t=4 seconds is valid. This result aligns with vertical motion equations and confirms how gravity affects falling objects from rest at a given height.

Solutions of a quadratic from a graph are the x-intercepts. The graph crosses the x-axis at x=−4 and x=0.8. These points indicate where y equals zero. Reading intercepts visually avoids algebraic solving and confirms roots directly from the graph’s intersection with the horizontal axis, ensuring accuracy in graphical interpretation.

The quadratic A(x)=84x−2x² opens downward, so maximum occurs at its vertex. Using x=−b/2a gives x=21. Substituting into width gives 84−42=42. Thus dimensions are 21 by 42. This configuration maximizes area due to symmetry around the vertex of the parabola representing area.

Maximum height occurs at the vertex of y=40x−5x². Using x=−b/2a gives x=4. Substituting yields y=40(4)−5(16)=80. This value represents the peak of the projectile’s motion. The quadratic’s negative leading coefficient confirms this point is a maximum rather than a minimum.

Parallel lines have equal slopes. The given line y=5x−4 has slope 5. Rewriting 5x−y+3=0 gives y=5x+3, which also has slope 5. Equal slopes confirm parallelism. Other options have different slopes, so they do not satisfy the required geometric condition of parallel lines.

Gradient equals change in y divided by change in x. For points (1,−3) and (3,7), change in y is 10 and change in x is 2. Dividing gives slope 5. This positive slope indicates a rising line and confirms consistent linear increase between the two points.

Solving simultaneously, subtracting equations eliminates y. From 2x+y=9 and 3x+2y=13, multiply first by two to get 4x+2y=18. Subtracting gives x=5. Substituting back gives y=−1. The ordered pair satisfies both equations, confirming the solution set algebraically.