What this deck covers
This deck focuses on Quadratic Equations, giving you a quick way to review the definitions, rules, and examples that matter most for PSAT Math.
Study Quadratic Equations in PSAT Math with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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What is the discriminant for ax2+bx+c=0?
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b2−4ac. The expression under the square root in the quadratic formula.
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This deck focuses on Quadratic Equations, giving you a quick way to review the definitions, rules, and examples that matter most for PSAT Math.
Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.
Answer: b2−4ac. The expression under the square root in the quadratic formula.
Answer: (h,k). The vertex is the point where the parabola reaches its extreme value.
Answer: 24. Δ=(−6)2−4(3)(1)=36−12=24.
Answer: r1r2=ac. Another application of Vieta's formulas.
Answer: y=a(x−r1)(x−r2). Each factor equals zero when x equals a root.
Answer: −2ab. The x-value where the parabola reaches its maximum or minimum.
Answer: b2−4ac<0. Negative discriminant means no real square root exists.
Answer: b2−4ac>0. Positive discriminant means the square root yields two different values.
Answer: x=2a−b±b2−4ac. Derived by completing the square on the general quadratic equation.
Answer: (0,c). Found by substituting x=0 into the equation.
Answer: x=2. Use x=−2ab=−2(2)−8=48=2.
Answer: y=a(x−h)2+k. Shows the vertex at (h,k) and vertical stretch/compression by a.
Answer: Opens downward. Negative leading coefficient creates an inverted U-shape.
Answer: x=2 and x=3. Factor as (x−2)(x−3)=0, so x=2 or x=3.
Answer: r1+r2=−ab. From Vieta's formulas for quadratic equations.
Answer: (0,c). Found by setting x=0 in the equation.
Answer: b2−4ac=0. Zero discriminant makes ±0=0, giving one repeated solution.
Answer: No real solutions (two complex solutions). Negative under square root means no real solutions exist.
Answer: Δ=b2−4ac. The expression under the square root in the quadratic formula.
Answer: Opens upward. Positive leading coefficient creates a U-shaped parabola.
Answer: Opens upward. Positive leading coefficient creates a U-shaped parabola.
Answer: Opens downward. Negative leading coefficient creates an inverted U-shape.
Answer: x=21 or x=−2. Factor as (2x−1)(x+2)=0, giving these solutions.
Answer: x=2. Use x=−2ab=−2(2)−8=48=2.
Answer: x=2 or x=3. Factor as (x−2)(x−3)=0, so x=2 or x=3.
Answer: ax2+bx+c=0, where a=0. Must have a=0 to ensure the equation is quadratic, not linear.
Answer: y=a(x−r1)(x−r2). Each factor equals zero when x equals a root.
Answer: x=−2ab. Found by completing the square or using calculus to find the minimum/maximum.
Answer: y=a(x−h)2+k. Shows the vertex at (h,k) and vertical stretch/compression by a.
Answer: x=21 and x=−2. Factor as (2x−1)(x+2)=0, so x=21 or x=−2.
Answer: One real solution (a repeated root). Zero discriminant makes ±0=0, giving one solution.
Answer: x=2a−b±b2−4ac. Derived by completing the square on the general quadratic equation.
Answer: 4. Calculate b2−4ac=(−4)2−4(3)(1)=16−12=4.
Answer: Two distinct real solutions. Positive discriminant means the square root yields two different values.
Answer: (3,−5). In vertex form y=a(x−h)2+k, the vertex is (h,k).
Answer: ax2+bx+c=0 with a=0. Must have a=0 to ensure the equation is quadratic, not linear.