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Algebra 2 Flashcards: Solve Quadratics By Multiple Methods

Study Solve Quadratics By Multiple Methods in Algebra 2 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 this deck covers

This deck focuses on Solve Quadratics By Multiple Methods, giving you a quick way to review the definitions, rules, and examples that matter most for Algebra 2.

How to use these flashcards

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.

Algebra 2 Flashcards: Solve Quadratics By Multiple Methods

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QUESTION

What does Δ<0\Delta<0Δ<0 guarantee about solutions of ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0?

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ANSWER

Two complex conjugate solutions. Negative discriminant means the parabola doesn't cross the x-axis.

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Flashcard 1: What does Δ<0\Delta<0Δ<0 guarantee about solutions of ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0?

Answer: Two complex conjugate solutions. Negative discriminant means the parabola doesn't cross the x-axis.

Flashcard 2: Solve by factoring: x2−4x+3=0x^2-4x+3=0x2−4x+3=0.

Answer: x=1x=1x=1 or x=3x=3x=3. Factor: (x−1)(x−3)=0(x-1)(x-3)=0(x−1)(x−3)=0, then use zero-product property.

Flashcard 3: What is Δ\DeltaΔ for 2x2−3x−2=02x^2-3x-2=02x2−3x−2=0?

Answer: Δ=25\Delta=25Δ=25. Δ=(−3)2−4(2)(−2)=9+16=25\Delta=(-3)^2-4(2)(-2)=9+16=25Δ=(−3)2−4(2)(−2)=9+16=25.

Flashcard 4: Solve using the quadratic formula: 2x2+4x+5=02x^2+4x+5=02x2+4x+5=0.

Answer: x=−1±62ix=-1\pm\frac{\sqrt{6}}{2}ix=−1±26​​i. Use a=2a=2a=2, b=4b=4b=4, c=5c=5c=5; −24=26i\sqrt{-24}=2\sqrt{6}i−24​=26​i.

Flashcard 5: Solve using the quadratic formula: x2+2x+5=0x^2+2x+5=0x2+2x+5=0.

Answer: x=−1±2ix=-1\pm 2ix=−1±2i. Negative discriminant gives complex solutions: −16=4i\sqrt{-16}=4i−16​=4i.

Flashcard 6: Solve using the quadratic formula: x2−2x+5=0x^2-2x+5=0x2−2x+5=0.

Answer: x=1±2ix=1\pm 2ix=1±2i. Negative discriminant gives complex solutions: −16=4i\sqrt{-16}=4i−16​=4i.

Flashcard 7: Solve using the quadratic formula: 3x2−12x+12=03x^2-12x+12=03x2−12x+12=0.

Answer: x=2x=2x=2. Use a=3a=3a=3, b=−12b=-12b=−12, c=12c=12c=12; discriminant equals 0.

Flashcard 8: Solve using the quadratic formula: 2x2+3x−2=02x^2+3x-2=02x2+3x−2=0.

Answer: x=12x=\frac{1}{2}x=21​ or x=−2x=-2x=−2. Use a=2a=2a=2, b=3b=3b=3, c=−2c=-2c=−2 in the quadratic formula.

Flashcard 9: Solve using the quadratic formula: x2+4x+1=0x^2+4x+1=0x2+4x+1=0.

Answer: x=−2±3x=-2\pm\sqrt{3}x=−2±3​. Use a=1a=1a=1, b=4b=4b=4, c=1c=1c=1 in the quadratic formula.

Flashcard 10: Solve using the quadratic formula: x2+5x+6=0x^2+5x+6=0x2+5x+6=0.

Answer: x=−2x=-2x=−2 or x=−3x=-3x=−3. Use a=1a=1a=1, b=5b=5b=5, c=6c=6c=6 in the quadratic formula.

Flashcard 11: Solve by completing the square: x2+2x−8=0x^2+2x-8=0x2+2x−8=0.

Answer: x=2x=2x=2 or x=−4x=-4x=−4. Complete the square: (x+1)2=9(x+1)^2=9(x+1)2=9, so x+1=±3x+1=\pm 3x+1=±3.

Flashcard 12: State the quadratic formula for solutions of ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0.

Answer: x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}x=2a−b±b2−4ac​​. Standard form for finding roots of any quadratic equation.

Flashcard 13: What is the discriminant for ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0?

Answer: Δ=b2−4ac\Delta=b^2-4acΔ=b2−4ac. The discriminant determines the nature of quadratic solutions.

Flashcard 14: Solve by completing the square: x2−4x−1=0x^2-4x-1=0x2−4x−1=0.

Answer: x=2±5x=2\pm\sqrt{5}x=2±5​. Complete the square: (x−2)2=5(x-2)^2=5(x−2)2=5, so x−2=±5x-2=\pm\sqrt{5}x−2=±5​.

Flashcard 15: What does Δ>0\Delta>0Δ>0 guarantee about solutions of ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0?

Answer: Two distinct real solutions. Positive discriminant means the parabola crosses the x-axis twice.

Flashcard 16: What does Δ=0\Delta=0Δ=0 guarantee about solutions of ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0?

Answer: One real double solution. Zero discriminant means the parabola touches the x-axis at one point.

Flashcard 17: What does Δ<0\Delta<0Δ<0 guarantee about solutions of ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0?

Answer: Two complex conjugate solutions. Negative discriminant means the parabola doesn't cross the x-axis.

Flashcard 18: State the square root property used to solve (x−h)2=k(x-h)^2=k(x−h)2=k.

Answer: (x−h)2=k⇒x−h=±k(x-h)^2=k\Rightarrow x-h=\pm\sqrt{k}(x−h)2=k⇒x−h=±k​. Take the square root of both sides to isolate x−hx-hx−h.

Flashcard 19: What are the solutions of x2=49x^2=49x2=49?

Answer: x=±7x=\pm 7x=±7. Take the square root of both sides: 49=7\sqrt{49}=749​=7.

Flashcard 20: What are the solutions of (x−3)2=16(x-3)^2=16(x−3)2=16?

Answer: x=7x=7x=7 or x=−1x=-1x=−1. Square root both sides: x−3=±4x-3=\pm 4x−3=±4, so x=3±4x=3\pm 4x=3±4.

Flashcard 21: What are the solutions of (x+5)2=9(x+5)^2=9(x+5)2=9?

Answer: x=−2x=-2x=−2 or x=−8x=-8x=−8. Square root both sides: x+5=±3x+5=\pm 3x+5=±3, so x=−5±3x=-5\pm 3x=−5±3.

Flashcard 22: What are the solutions of 2(x−1)2=182(x-1)^2=182(x−1)2=18?

Answer: x=4x=4x=4 or x=−2x=-2x=−2. Divide by 2 first: (x−1)2=9(x-1)^2=9(x−1)2=9, then x−1=±3x-1=\pm 3x−1=±3.

Flashcard 23: What are the solutions of (x−4)2=0(x-4)^2=0(x−4)2=0?

Answer: x=4x=4x=4. Only one solution since the square root of 0 is 0.

Flashcard 24: What are the solutions of x2+8x+16=0x^2+8x+16=0x2+8x+16=0?

Answer: x=−4x=-4x=−4. Perfect square trinomial: (x+4)2=0(x+4)^2=0(x+4)2=0, so x=−4x=-4x=−4.

Flashcard 25: Factor and solve x2−9=0x^2-9=0x2−9=0.

Answer: x=±3x=\pm 3x=±3. Difference of squares: (x−3)(x+3)=0(x-3)(x+3)=0(x−3)(x+3)=0.

Flashcard 26: Factor and solve x2−5x=0x^2-5x=0x2−5x=0.

Answer: x=0x=0x=0 or x=5x=5x=5. Factor out xxx: x(x−5)=0x(x-5)=0x(x−5)=0, then use zero-product property.

Flashcard 27: Factor and solve x2+7x+12=0x^2+7x+12=0x2+7x+12=0.

Answer: x=−3x=-3x=−3 or x=−4x=-4x=−4. Factor: (x+3)(x+4)=0(x+3)(x+4)=0(x+3)(x+4)=0, then use zero-product property.

Flashcard 28: Factor and solve x2−6x+8=0x^2-6x+8=0x2−6x+8=0.

Answer: x=2x=2x=2 or x=4x=4x=4. Factor: (x−2)(x−4)=0(x-2)(x-4)=0(x−2)(x−4)=0, then use zero-product property.

Flashcard 29: Factor and solve x2+x−6=0x^2+x-6=0x2+x−6=0.

Answer: x=2x=2x=2 or x=−3x=-3x=−3. Factor: (x+3)(x−2)=0(x+3)(x-2)=0(x+3)(x−2)=0, then use zero-product property.

Flashcard 30: Solve by factoring: 2x2−8x=02x^2-8x=02x2−8x=0.

Answer: x=0x=0x=0 or x=4x=4x=4. Factor out 2x2x2x: 2x(x−4)=02x(x-4)=02x(x−4)=0, then use zero-product property.