Math: Quadratic Formula Tune Practice
Practice using the quadratic formula to solve quadratic equations
Math: Quadratic Formula Tune Practice
Practice using the quadratic formula to solve quadratic equations
Math - Grade 9-12
- 1
Solve x^2 + 5x + 6 = 0 using the quadratic formula.
Start by finding the discriminant b^2 - 4ac.
Here a = 1, b = 5, and c = 6. The quadratic formula gives x = (-5 ± sqrt(25 - 24)) / 2 = (-5 ± 1) / 2, so the solutions are x = -2 and x = -3. - 2
Solve 2x^2 - 3x - 2 = 0 using the quadratic formula.
Here a = 2, b = -3, and c = -2. The formula gives x = (3 ± sqrt(9 + 16)) / 4 = (3 ± 5) / 4, so the solutions are x = 2 and x = -1/2. - 3
Solve x^2 - 4x + 7 = 0 using the quadratic formula. State whether the solutions are real or complex.
A negative discriminant means the equation has two complex solutions.
Here a = 1, b = -4, and c = 7. The formula gives x = (4 ± sqrt(16 - 28)) / 2 = (4 ± sqrt(-12)) / 2 = 2 ± i sqrt(3), so the solutions are complex. - 4
For the equation 3x^2 + 6x + 3 = 0, use the discriminant to predict the number of real solutions. Then solve.
When the discriminant is 0, the parabola touches the x-axis once.
Here a = 3, b = 6, and c = 3. The discriminant is 6^2 - 4(3)(3) = 36 - 36 = 0, so there is one real solution. The solution is x = -6 / 6 = -1. - 5
Solve 4x^2 + 4x - 3 = 0 using the quadratic formula.
Here a = 4, b = 4, and c = -3. The formula gives x = (-4 ± sqrt(16 + 48)) / 8 = (-4 ± 8) / 8, so the solutions are x = 1/2 and x = -3/2. - 6
Solve 5x^2 - 2x + 1 = 0 using the quadratic formula. Write the answer in simplest complex form.
Remember that sqrt(-16) = 4i.
Here a = 5, b = -2, and c = 1. The formula gives x = (2 ± sqrt(4 - 20)) / 10 = (2 ± sqrt(-16)) / 10 = (2 ± 4i) / 10, so x = (1 ± 2i) / 5. - 7
A student is solving 2x^2 + 7x + 3 = 0 and writes x = (-7 ± sqrt(7^2 - 4(2)(3))) / 2. Explain the mistake and write the correct setup.
Check the denominator of the quadratic formula carefully.
The mistake is that the denominator should be 2a, not just 2. Since a = 2, the correct setup is x = (-7 ± sqrt(7^2 - 4(2)(3))) / 4. - 8
Solve 2x^2 + 7x + 3 = 0 using the correct quadratic formula setup.
Here a = 2, b = 7, and c = 3. The formula gives x = (-7 ± sqrt(49 - 24)) / 4 = (-7 ± 5) / 4, so the solutions are x = -1/2 and x = -3. - 9
The graph of a quadratic equation crosses the x-axis at x = 1 and x = 4. What are the solutions of the equation, and what does this mean about the discriminant?
The x-intercepts of a parabola are the real solutions of the related equation.
The solutions are x = 1 and x = 4 because the x-intercepts are the zeros of the quadratic equation. Since there are two real solutions, the discriminant is positive. - 10
Solve -x^2 + 6x - 8 = 0 using the quadratic formula.
Here a = -1, b = 6, and c = -8. The formula gives x = (-6 ± sqrt(36 - 32)) / -2 = (-6 ± 2) / -2, so the solutions are x = 2 and x = 4. - 11
Find the exact solutions of x^2 + 2x - 5 = 0 using the quadratic formula.
Simplify sqrt(24) before reducing the fraction.
Here a = 1, b = 2, and c = -5. The formula gives x = (-2 ± sqrt(4 + 20)) / 2 = (-2 ± sqrt(24)) / 2 = (-2 ± 2sqrt(6)) / 2, so x = -1 ± sqrt(6). - 12
A ball's height in meters is modeled by h = -5t^2 + 20t + 1. Use the quadratic formula to find when the ball hits the ground. Round your answer to the nearest hundredth.
Only the positive time makes sense in this situation.
Set h = 0 to get -5t^2 + 20t + 1 = 0. Using a = -5, b = 20, and c = 1 gives t = (-20 ± sqrt(400 + 20)) / -10 = (-20 ± sqrt(420)) / -10. The positive time is about 4.05 seconds, so the ball hits the ground after about 4.05 seconds.