Algebra 1 Section 12.2

Solving Quadratic Equations Another method of solving equations of the form x2 – c = 0 is to isolate the squared term and then take the square root of each side of the equation.

Example 1 Solve x2 – 64 = 0. x2 = 64 x2 = 64 |x| = 8 x = ±8

Square Root Property If n ≥ 0 and x2 = n, then x = ±n. If n > 0, then x has two real solutions. If n = 0, then x has one real solution.

Square Root Property If n < 0 and x2 = n, then x has no real solutions.

Example 2 Solve 2x2 – 24 = 0. 2x2 = 24 x2 = 12 x = ± 12 x = ± 2 3

Example 3 Solve (x + 2)2 = 36. x + 2 = ± 6 x = -2 ± 6 x = 4, -8 Did you check each solution in the original equation?

Solving Equations of the Form ax2 – c = 0 Isolate the squared expression. Apply the Square Root Property. Simplify the radical and solve each equation.

Example 4 Solve (x – 5)2 = 21. x – 5 = ± 21 x = 5 ± 21 x ≈ 5 ± 4.58 The solution set is {0.42, 9.58}.

Example 5 Solve 7x2 + 6 = 2x2 + 8. 5x2 + 6 = 8 5x2 = 2 x2 = x = ± 2 5

Example 5 x = ± 2 5 x = ± 2 5 5 • = ± 10 5

Solving Quadratic Equations Solving x2 + 7 = 0 yields the solutions x = ± -7. The square root of -7 is not a real number.

Solving Quadratic Equations For now, solve the problem, leaving the negative number under the radical; and state that there are no real solutions.

The Free Fall Equation Assuming negligible air resistance, the height of a dropped object can be modeled by the function h(t) = -16t2 + hi, where t is the time in seconds and hi is the initial height in feet.

Example 6 h(t) = -16t2 + hi 0 = -16t2 + 150 16t2 = 150 t2 = t = ± 150

Only t ≈ 3.06 sec is a reasonable answer. Example 6 t = ± 150 16 t = ± 5 6 4 ≈ ± 3.06 Only t ≈ 3.06 sec is a reasonable answer.

Homework: pp. 491-493