Quadratic Equations Graphing & Factoring Quiz

1) What are the solutions to x² - 7x + 12 = 0 by factoring?

X = 2, x = 6

X = 3, x = 4

X = 1, x = 12

X = -3, x = -4

Factoring x² - 7x + 12, we find two numbers that multiply to 12 and add to -7: -3 and -4. Therefore, the factored form is (x - 3)(x - 4) = 0. Setting each factor equal to zero gives the solutions x = 3 and x = 4. The discriminant confirms two real roots. Verification by substitution shows that both solutions satisfy the original equation.

Explanation

Factoring x² - 7x + 12, we find two numbers that multiply to 12 and add to -7: -3 and -4. Therefore, the factored form is (x - 3)(x - 4) = 0. Setting each factor equal to zero gives the solutions x = 3 and x = 4. The discriminant confirms two real roots. Verification by substitution shows that both solutions satisfy the original equation.

2) Solve 2x² + 5x - 3 = 0 by factoring.

X = 1, x = -3/2

X = -3, x = 1/2

X = -1, x = 3/2

X = 3, x = -1/2

To solve 2x² + 5x - 3 = 0 by factoring, we use the AC method. The factors of 2 * -3 = -6 that add to 5 are 6 and -1. The equation becomes 2x² + 6x - x - 3 = 0, which factors as (2x - 1)(x + 3) = 0. Solving each factor gives x = 1/2 and x = -3, confirming the solutions.

Explanation

To solve 2x² + 5x - 3 = 0 by factoring, we use the AC method. The factors of 2 * -3 = -6 that add to 5 are 6 and -1. The equation becomes 2x² + 6x - x - 3 = 0, which factors as (2x - 1)(x + 3) = 0. Solving each factor gives x = 1/2 and x = -3, confirming the solutions.

3) The graph of y = x² - 4x - 5 crosses the x-axis at which points?

(0, 0) and (4, 0)

(-1, 0) and (5, 0)

No real x-intercepts

(1, 0) and (-5, 0)

The graph of y = x² - 4x - 5 crosses the x-axis where y = 0. Factoring the quadratic gives (x - 5)(x + 1) = 0, which gives the solutions x = 5 and x = -1. These are the x-intercepts where the parabola crosses the x-axis. The discriminant confirms two real roots, and the parabola opens upward.

Explanation

The graph of y = x² - 4x - 5 crosses the x-axis where y = 0. Factoring the quadratic gives (x - 5)(x + 1) = 0, which gives the solutions x = 5 and x = -1. These are the x-intercepts where the parabola crosses the x-axis. The discriminant confirms two real roots, and the parabola opens upward.

4) Factor x² + 8x + 16 completely.

(x + 4)(x + 4)

(x + 4)²

(x - 4)²

(x + 2)(x + 8)

The quadratic x² + 8x + 16 is a perfect square trinomial. It factors as (x + 4)², because (4)² = 16 and 8/2 = 4. The equation (x + 4)² = 0 has a repeated root at x = -4, meaning the vertex of the parabola touches the x-axis. The discriminant is 0, confirming only one real solution.

Explanation

The quadratic x² + 8x + 16 is a perfect square trinomial. It factors as (x + 4)², because (4)² = 16 and 8/2 = 4. The equation (x + 4)² = 0 has a repeated root at x = -4, meaning the vertex of the parabola touches the x-axis. The discriminant is 0, confirming only one real solution.

5) What are the zeros of the quadratic graphed with vertex at (3, -4) and x-intercepts at 1 and 5?

X = 1 and x = 5

X = 3 only

X = 0 and x = 3

X = -1 and x = -5

The zeros of the quadratic function are the x-intercepts, given as 1 and 5. These are the points where the parabola crosses the x-axis. The vertex is at (3, -4), which is the midpoint between the intercepts, confirming that the axis of symmetry is at x = 3. The equation can be factored as (x - 1)(x - 5) = 0.

Explanation

The zeros of the quadratic function are the x-intercepts, given as 1 and 5. These are the points where the parabola crosses the x-axis. The vertex is at (3, -4), which is the midpoint between the intercepts, confirming that the axis of symmetry is at x = 3. The equation can be factored as (x - 1)(x - 5) = 0.

6) Solve by factoring: 3x² - 12x = 0.

X = 0 only

X = 0, x = -4

X = 0, x = 4

X = 3, x = 4

The quadratic 3x² - 12x = 0 factors as 3x(x - 4) = 0. Setting each factor equal to zero gives x = 0 and x = 4 as the solutions. The discriminant confirms two real roots. The solution x = 0 is also the y-intercept, where the parabola touches the y-axis. Verification by substitution shows that both solutions satisfy the original equation.

Explanation

The quadratic 3x² - 12x = 0 factors as 3x(x - 4) = 0. Setting each factor equal to zero gives x = 0 and x = 4 as the solutions. The discriminant confirms two real roots. The solution x = 0 is also the y-intercept, where the parabola touches the y-axis. Verification by substitution shows that both solutions satisfy the original equation.

7) Which equation has solutions x = -5 and x = 2 after factoring?

X² + 7x - 10 = 0

X² + 3x - 10 = 0

X² - 3x - 10 = 0

X² - 7x + 10 = 0

The equation x² + 3x - 10 = 0 has solutions x = -5 and x = 2. Factoring this quadratic gives (x + 5)(x - 2) = 0, which gives the solutions x = -5 and x = 2. Expanding confirms the factorization. The discriminant confirms two real roots. The equation matches the sum and product of the roots: -b/a = -3.

Explanation

The equation x² + 3x - 10 = 0 has solutions x = -5 and x = 2. Factoring this quadratic gives (x + 5)(x - 2) = 0, which gives the solutions x = -5 and x = 2. Expanding confirms the factorization. The discriminant confirms two real roots. The equation matches the sum and product of the roots: -b/a = -3.

8) From a graph opening upward with x-intercepts at -2 and 3, the equation could be:

(x - 2)(x - 3) = 0

(x + 2)(x + 3) = 0

(x + 2)(x - 3) = 0

(x - 2)(x + 3) = 0

The graph has x-intercepts at -2 and 3, so the equation can be factored as (x + 2)(x - 3) = 0. Expanding this gives x² - x - 6 = 0, confirming the equation. Since the parabola opens upward, the positive leading coefficient confirms the upward direction. Other choices lead to incorrect x-intercepts. The solution correctly represents the graph's intercepts.

Explanation

The graph has x-intercepts at -2 and 3, so the equation can be factored as (x + 2)(x - 3) = 0. Expanding this gives x² - x - 6 = 0, confirming the equation. Since the parabola opens upward, the positive leading coefficient confirms the upward direction. Other choices lead to incorrect x-intercepts. The solution correctly represents the graph's intercepts.

9) Solve 4x² - 9 = 0 by factoring (difference of squares).

X = 9/4, x = -9/4

X = 2, x = -2

X = 3, x = -3

X = 3/2, x = -3/2

The equation 4x² - 9 = 0 is a difference of squares. It factors as (2x - 3)(2x + 3) = 0, and solving each factor gives x = 3/2 and x = -3/2. The discriminant confirms two real roots, and no middle term simplifies the process. The equation demonstrates the use of the difference of squares to find solutions.

Explanation

The equation 4x² - 9 = 0 is a difference of squares. It factors as (2x - 3)(2x + 3) = 0, and solving each factor gives x = 3/2 and x = -3/2. The discriminant confirms two real roots, and no middle term simplifies the process. The equation demonstrates the use of the difference of squares to find solutions.

10) The zeros of y = -x² + 6x - 8 are found by graphing. They are:

X = 0 and x = 6

X = 1 and x = 8

X = 2 and x = 4

X = -2 and x = -4

To solve -x² + 6x - 8 = 0, we first multiply the entire equation by -1 to simplify to x² - 6x + 8 = 0. Factoring this gives (x - 2)(x - 4) = 0, and solving for x gives the zeros x = 2 and x = 4. The discriminant confirms two distinct real roots, and the symmetry about the axis x = 3 is confirmed.

Explanation

To solve -x² + 6x - 8 = 0, we first multiply the entire equation by -1 to simplify to x² - 6x + 8 = 0. Factoring this gives (x - 2)(x - 4) = 0, and solving for x gives the zeros x = 2 and x = 4. The discriminant confirms two distinct real roots, and the symmetry about the axis x = 3 is confirmed.

11) Factor and solve: x² - 5x - 14 = 0.

X = 14, x = -1

X = 7, x = -2

X = -7, x = 2

X = 5, x = -14/5

The quadratic equation x² - 5x - 14 = 0 factors as (x - 7)(x + 2) = 0, giving the solutions x = 7 and x = -2. Verification through substitution shows that both solutions satisfy the original equation. The discriminant confirms two real roots, and the negative constant indicates roots with opposite signs.

Explanation

The quadratic equation x² - 5x - 14 = 0 factors as (x - 7)(x + 2) = 0, giving the solutions x = 7 and x = -2. Verification through substitution shows that both solutions satisfy the original equation. The discriminant confirms two real roots, and the negative constant indicates roots with opposite signs.

12) A parabola touches the x-axis at x = -3 only. The factored equation is:

(x + 3)(x - 3) = 0

(x - 3)² = 0

(x + 3)² = 0

X(x + 3) = 0

The parabola touches the x-axis at x = -3, which indicates a double root. The factored form is (x + 3)² = 0. The discriminant is 0, confirming only one real, repeated root. This corresponds to a vertex at x = -3, where the graph just touches the x-axis and does not cross it. Other forms do not fit this condition.

Explanation

The parabola touches the x-axis at x = -3, which indicates a double root. The factored form is (x + 3)² = 0. The discriminant is 0, confirming only one real, repeated root. This corresponds to a vertex at x = -3, where the graph just touches the x-axis and does not cross it. Other forms do not fit this condition.