Download 9-5 Solving Quadratic Equations by Using the Quadratic Formula

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Two-body problem in general relativity wikipedia , lookup

Two-body Dirac equations wikipedia , lookup

Euler equations (fluid dynamics) wikipedia , lookup

Itô diffusion wikipedia , lookup

Differential equation wikipedia , lookup

Debye–Hückel equation wikipedia , lookup

Exact solutions in general relativity wikipedia , lookup

Partial differential equation wikipedia , lookup

Schwarzschild geodesics wikipedia , lookup

Transcript
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
9-5 Solving Quadratic Equations by Using the Quadratic Formula
To solve the standard form of the quadratic equation, ax2 + bx + c = 0, use the Quadratic Formula.
The solutions of ax2 + bx + c = 0, where a ≠ 0, are given by
Quadratic Formula
b  b 2  4ac
x=
2a
Example 1: Solve x2 +2x = 3by using the
Quadratic Formula.
b  b2  4ac
or x =
2a

Example 2: Solve x2 – 6x – 2 = 0
by using the Quadratic Formula. Round to the
nearest tenth
if necessary.
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Exercises
Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.
1.
x2 – 3x + 2 = 0
2.
16x2 – 8x = -1
3.
3x2 + 2x = 8
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
9-5 Solving Quadratic Equations by Using the Quadratic Formula
The Discriminant In the Quadratic Formula, the expression under the radical sign, b2 – 4ac, is called the discriminant.
The discriminant can be used to determine the number of real solutions for a quadratic equation.
Case 1: b2 – 4ac < 0
no real solutions
Case 2: b2 – 4ac = 0
one real solution
Case 3: b2 – 4ac > 0
two real solutions
Example: State the value of the discriminant for each equation. Then determine the number of real solutions of
the equation.
a.
12x2 + 5x = 4
b.
2x2 + 3x = -4
Exercises
State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.
1.
3x2 + 2x – 3 = 0
discriminant = ______________
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Exercises
State the value of the discriminant for each equation. Then determine the number of real solutions of the equation.
2.
2x2 – 10x – 9 = 0
discriminant = _______________
3.
6x2 – 10x + 10 = 0
discriminant = ________________
4.
9 – 18x + 9x2 = 0
discriminat = __________________
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
5.
16x2 + 16x + 4 = 0
discriminant = __________________
6.
3x2 – 18x = -14
discriminant = ____________________
7.
4x2 - 4x + 4 = 3
discriminant = ____________________