The Quadratic Formula For any quadratic equation of

  • Slides: 13
Download presentation

The Quadratic Formula For any quadratic equation of the form The solutions are given

The Quadratic Formula For any quadratic equation of the form The solutions are given by the formula:

Example Solve 3 x 2 + 5 x = 2 using the quadratic formula.

Example Solve 3 x 2 + 5 x = 2 using the quadratic formula. Solution First determine a, b, and c: 3 x 2 + 5 x – 2 = 0; a = 3, b = 5, and c = – 2. Substituting

The solutions are 1/3 and – 2.

The solutions are 1/3 and – 2.

Example Solve x 2 + 7 = 2 x using the quadratic formula. Solution

Example Solve x 2 + 7 = 2 x using the quadratic formula. Solution First determine a, b, and c: x 2 – 2 x + 7 = 0; a = 1, b = – 2, and c = 7. Substituting

The solutions are

The solutions are

Examples Solve the following quadratic equations:

Examples Solve the following quadratic equations:

Discriminant The radicand in the quadratic formula is called the discriminant. b 2 –

Discriminant The radicand in the quadratic formula is called the discriminant. b 2 – 4 ac • If the discriminant is positive, then there will be two real solutions • If the discriminant is 0, then there will be one real solution. • If the discriminant is negative, then there will be two imaginary solutions.

Solving Quadratic Equations Method 1: Factoring Method 2: Using Square Root Property Method 3:

Solving Quadratic Equations Method 1: Factoring Method 2: Using Square Root Property Method 3: Completing the Square Method 4: Using Quadratic Formula

Writing Equations from Solutions We know by the principle of zero products that (x

Writing Equations from Solutions We know by the principle of zero products that (x – 1)(x + 4) = 0 has solutions 1 and -4. If we know the solutions of an equation, we can write an equation, using the principle in reverse.

Example Find an equation for which 5 and – 4/3 are solutions. Solution x

Example Find an equation for which 5 and – 4/3 are solutions. Solution x = 5 or x = – 4/3 x – 5 = 0 or x + 4/3 = 0 (x – 5)(x + 4/3) = 0 x 2 – 5 x + 4/3 x – 20/3 = 0 3 x 2 – 11 x – 20 = 0 Get 0’s on one side Using the principle of zero products Multiplying Combining like terms and clearing fractions

Example Find an equation for which 3 i and – 3 i are solutions.

Example Find an equation for which 3 i and – 3 i are solutions. Solution x = 3 i or x = – 3 i x – 3 i = 0 or x + 3 i = 0 (x – 3 i)(x + 3 i) = 0 x 2 – 3 ix + 3 ix – 9 i 2 = 0 x 2 + 9 = 0 Get 0’s on one side Using the principle of zero products Multiplying Combining like terms

Example The braking distance, d (in feet), of a car going v miles per

Example The braking distance, d (in feet), of a car going v miles per hour is given by How fast would a car be traveling if its braking distance is 150 ft?