Solving Problems Involving Quadratics

1) Solve using factoring: 3x² − 13x − 10 = 0

X = −2/3, 5

X = 2/3, −5

X = −5, −2/3

X = 3/2, 5

Factoring 3x² − 13x − 10 requires identifying two numbers that multiply to −30 and add to −13. These numbers are −15 and 2. Splitting the middle term gives 3x² − 15x + 2x − 10. Factoring by grouping results in (3x + 2)(x − 5) = 0. Solving each factor gives x = −2/3 and x = 5 as the solutions.

Explanation

Factoring 3x² − 13x − 10 requires identifying two numbers that multiply to −30 and add to −13. These numbers are −15 and 2. Splitting the middle term gives 3x² − 15x + 2x − 10. Factoring by grouping results in (3x + 2)(x − 5) = 0. Solving each factor gives x = −2/3 and x = 5 as the solutions.

2) For P(x) = −2x² + 13x − 15, when does profit equal zero?

1500 and 5000

150 and 500

1.5 and 5

1500 and 50000

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 skateboards, multiply both values by 1000. This gives 1500 and 5000 skateboards, which correctly represent the break-even quantities.

Explanation

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 skateboards, multiply both values by 1000. This gives 1500 and 5000 skateboards, which correctly represent the break-even quantities.

3) When does h(t) = −5t² + 35t hit the ground?

1

3.5

7

61.25

The ball hits the ground when height equals zero. Setting −5t² + 35t = 0 and factoring out t gives t(−5t + 35) = 0. The solutions are t = 0 and t = 7. Time zero represents launch, not landing. Therefore, the ball hits the ground at 7 seconds, which is the physically meaningful solution.

Explanation

The ball hits the ground when height equals zero. Setting −5t² + 35t = 0 and factoring out t gives t(−5t + 35) = 0. The solutions are t = 0 and t = 7. Time zero represents launch, not landing. Therefore, the ball hits the ground at 7 seconds, which is the physically meaningful solution.

4) The roots of a quadratic graph occur where it crosses the

Y-axis

Vertex

X-axis

Axis of symmetry

Roots of a quadratic are the x-values where the graph intersects the x-axis. At these points, the y-value equals zero, meaning the equation is satisfied. The vertex represents the maximum or minimum value, not the solutions. Therefore, the x-axis intersection correctly identifies where the quadratic equals zero and where real solutions occur graphically.

Explanation

Roots of a quadratic are the x-values where the graph intersects the x-axis. At these points, the y-value equals zero, meaning the equation is satisfied. The vertex represents the maximum or minimum value, not the solutions. Therefore, the x-axis intersection correctly identifies where the quadratic equals zero and where real solutions occur graphically.

5) Solve −5t² + 80 = 0

−4

4

5

80

To find when the object hits the ground, set −5t² + 80 = 0. Rearranging gives 5t² = 80, so t² = 16. Taking the square root yields t = ±4. Since time cannot be negative in real-world motion, the valid solution is t = 4 seconds, which represents when the object reaches the ground.

Explanation

To find when the object hits the ground, set −5t² + 80 = 0. Rearranging gives 5t² = 80, so t² = 16. Taking the square root yields t = ±4. Since time cannot be negative in real-world motion, the valid solution is t = 4 seconds, which represents when the object reaches the ground.

6) Maximum area of A(x) = 84x − 2x² occurs when x equals

14

21

42

84

The function A(x) = 84x − 2x² is a downward-opening parabola. The maximum occurs at the vertex, found using x = −b/(2a). Here, a = −2 and b = 84. Substituting gives x = −84/(−4) = 21. This value produces the largest possible area according to quadratic optimization principles.

Explanation

The function A(x) = 84x − 2x² is a downward-opening parabola. The maximum occurs at the vertex, found using x = −b/(2a). Here, a = −2 and b = 84. Substituting gives x = −84/(−4) = 21. This value produces the largest possible area according to quadratic optimization principles.

7) Maximum height of y = 40x − 5x² is

50

60

75

80

The maximum value of y = 40x − 5x² occurs at the vertex. Using x = −b/(2a) gives x = −40/(−10) = 4. Substituting x = 4 into the equation gives y = 160 − 80 = 80. Therefore, the maximum height of the stone is 80 meters, which occurs at four seconds after launch.

Explanation

The maximum value of y = 40x − 5x² occurs at the vertex. Using x = −b/(2a) gives x = −40/(−10) = 4. Substituting x = 4 into the equation gives y = 160 − 80 = 80. Therefore, the maximum height of the stone is 80 meters, which occurs at four seconds after launch.

8) Number of real solutions depends on the

Vertex

Discriminant

Axis of symmetry

Y-intercept

The discriminant, given by b² − 4ac, determines the number of real solutions of a quadratic equation. If it is positive, there are two real solutions. If zero, one real solution. If negative, no real solutions. Therefore, the discriminant alone fully determines whether real roots exist, making it the correct determining factor.

Explanation

The discriminant, given by b² − 4ac, determines the number of real solutions of a quadratic equation. If it is positive, there are two real solutions. If zero, one real solution. If negative, no real solutions. Therefore, the discriminant alone fully determines whether real roots exist, making it the correct determining factor.

9) A quadratic with no real roots has discriminant

Positive

Zero

Negative

Undefined

A quadratic equation has no real solutions when the discriminant is negative. This means b² − 4ac is less than zero, causing the square root term to be undefined in the real number system. As a result, the graph does not intersect the x-axis, and the equation has no real roots, only complex solutions.

Explanation

A quadratic equation has no real solutions when the discriminant is negative. This means b² − 4ac is less than zero, causing the square root term to be undefined in the real number system. As a result, the graph does not intersect the x-axis, and the equation has no real roots, only complex solutions.

10) The vertex of y = −2x² + 8x − 3 occurs at x =

−4

2

4

8

The x-coordinate of the vertex is found using x = −b/(2a). For y = −2x² + 8x − 3, a = −2 and b = 8. Substituting gives x = −8/(−4) = 2. This value identifies the axis of symmetry and the location of the maximum point of the parabola.

Explanation

The x-coordinate of the vertex is found using x = −b/(2a). For y = −2x² + 8x − 3, a = −2 and b = 8. Substituting gives x = −8/(−4) = 2. This value identifies the axis of symmetry and the location of the maximum point of the parabola.