Quadratic Equations
Solving and analyzing quadratic equations
Quadratic Equations
Solving and analyzing quadratic equations
Math - Grade 9-12
- 1
Solve the equation x^2 - 9 = 0.
Isolate x^2 first, then take the square root of both sides.
The solutions are x = 3 and x = -3 because x^2 = 9 and the square roots of 9 are 3 and -3. - 2
Solve the equation x^2 + 5x + 6 = 0 by factoring.
Find two numbers that multiply to 6 and add to 5.
The solutions are x = -2 and x = -3 because the equation factors as (x + 2)(x + 3) = 0. - 3
Solve the equation x^2 - 7x + 12 = 0 by factoring.
The solutions are x = 3 and x = 4 because the equation factors as (x - 3)(x - 4) = 0. - 4
Solve the equation 2x^2 - 8 = 0.
Divide both sides by 2 before taking square roots.
The solutions are x = 2 and x = -2 because 2x^2 = 8, so x^2 = 4, and the square roots of 4 are 2 and -2. - 5
Solve the equation x^2 + 2x - 8 = 0 by factoring.
The solutions are x = 2 and x = -4 because the equation factors as (x + 4)(x - 2) = 0. - 6
Solve the equation x^2 - 6x + 5 = 0 by factoring.
Find two numbers that multiply to 5 and add to -6.
The solutions are x = 1 and x = 5 because the equation factors as (x - 1)(x - 5) = 0. - 7
Solve the equation x^2 + 4x + 1 = 0 using the quadratic formula.
The solutions are x = -2 + square root of 3 and x = -2 - square root of 3. Using the quadratic formula gives x = (-4 plus or minus square root of 16 - 4) divided by 2, which simplifies to -2 plus or minus square root of 3. - 8
Solve the equation 3x^2 - 12x = 0 by factoring.
Factor out the greatest common factor first.
The solutions are x = 0 and x = 4 because the equation factors as 3x(x - 4) = 0. - 9
Solve the equation x^2 = 16.
The solutions are x = 4 and x = -4 because both 4 squared and negative 4 squared equal 16. - 10
Write the equation x^2 - 4x - 12 = 0 in factored form and then solve it.
Find two numbers that multiply to -12 and add to -4.
The factored form is (x - 6)(x + 2) = 0, so the solutions are x = 6 and x = -2. - 11
For the quadratic y = x^2 - 6x + 8, find the x-intercepts by solving x^2 - 6x + 8 = 0.
The x-intercepts are x = 2 and x = 4 because the equation factors as (x - 2)(x - 4) = 0. The intercept points are (2, 0) and (4, 0). - 12
Find the vertex of y = x^2 - 4x + 7.
Use x = -b divided by 2a to find the x-coordinate of the vertex.
The vertex is (2, 3). Completing the square gives y = (x - 2)^2 + 3, so the vertex is at x = 2 and y = 3. - 13
Solve the equation x^2 + 10x + 25 = 0.
The solution is x = -5. This is a repeated solution because the equation factors as (x + 5)^2 = 0. - 14
Determine the number of real solutions for the equation x^2 + 2x + 5 = 0.
Use the discriminant to decide how many real solutions there are.
The equation has no real solutions because the discriminant is b^2 - 4ac = 4 - 20 = -16, which is negative. - 15
A ball is launched upward and its height is modeled by h = -16t^2 + 48t. Find the times when the ball is on the ground.
The ball is on the ground at t = 0 seconds and t = 3 seconds because setting h = 0 gives -16t^2 + 48t = 0, which factors as -16t(t - 3) = 0.