10-4 Solving Quadratic Equations by Using the Quadratic Formula

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Presentation transcript:

10-4 Solving Quadratic Equations by Using the Quadratic Formula

Objectives Solve quadratic equations by using the Quadratic Formula. Determine the number of solutions of a quadratic equation by using the discriminant.

To add fractions, you need a common denominator. Remember!

Example 1A: Using the Quadratic Formula Solve using the Quadratic Formula. 6x2 + 5x – 4 = 0 6x2 + 5x + (–4) = 0 Identify a, b, and c. Use the Quadratic Formula. Substitute 6 for a, 5 for b, and –4 for c. Simplify.

6x2 + 5x – 4 = 0 Solve using the Quadratic Formula. Example 1A Continued Solve using the Quadratic Formula. 6x2 + 5x – 4 = 0 Simplify. Write as two equations. Solve each equation.

Example 1B: Using the Quadratic Formula Solve using the Quadratic Formula. x2 = x + 20 Write in standard form. Identify a, b, and c. 1x2 + (–1x) + (–20) = 0 Use the quadratic formula. Substitute 1 for a, –1 for b, and –20 for c. Simplify.

x = 5 or x = –4 Example 1B Continued Simplify. Write as two equations. Solve each equation.

You can graph the related quadratic function to see if your solutions are reasonable. Helpful Hint

–3x2 + 5x + 2 = 0 –3x2 + 5x + 2 = 0 Solve using the Quadratic Formula. Check It Out! Example 1a Solve using the Quadratic Formula. –3x2 + 5x + 2 = 0 –3x2 + 5x + 2 = 0 Identify a, b, and c. Use the Quadratic Formula. Substitute –3 for a, 5 for b, and 2 for c. Simplify

Check It Out! Example 1a Continued Simplify. Write as two equations. Solve each equation. or x = 2

Solve using the Quadratic Formula. Check It Out! Example 1b Solve using the Quadratic Formula. 2 – 5x2 = –9x –5x2 + 9x + 2 = 0 Write in standard form. Identify a, b, and c. Use the Quadratic Formula. Substitute –5 for a, 9 for b, and 2 for c. Simplify

Check It Out! Example 1b Continued Solve using the Quadratic Formula. Simplify. Write as two equations. Solve each equation. or x = 2

Many quadratic equations can be solved by graphing, factoring, taking the square root, or completing the square. Some cannot be solved by any of these methods, but you can always use the Quadratic Formula to solve any quadratic equation.

Example 2: Using the Quadratic Formula to Estimate Solutions Solve x2 + 3x – 7 = 0 using the Quadratic Formula. Check reasonableness Use a calculator: x ≈ 1.54 or x ≈ –4.54.

Use a calculator: x ≈ 3.87 or x ≈ 0.13. Check It Out! Example 2 Solve 2x2 – 8x + 1 = 0 using the Quadratic Formula. Check reasonableness Use a calculator: x ≈ 3.87 or x ≈ 0.13.

If the quadratic equation is in standard form, the discriminant of a quadratic equation is b2 – 4ac, the part of the equation under the radical sign. Recall that quadratic equations can have two, one, or no real solutions. You can determine the number of solutions of a quadratic equation by evaluating its discriminant.

Example 3: Using the Discriminant Find the number of solutions of each equation using the discriminant. A. B. C. 3x2 – 2x + 2 = 0 2x2 + 11x + 12 = 0 x2 + 8x + 16 = 0 a = 3, b = –2, c = 2 a = 2, b = 11, c = 12 a = 1, b = 8, c = 16 b2 – 4ac b2 – 4ac b2 – 4ac (–2)2 – 4(3)(2) 112 – 4(2)(12) 82 – 4(1)(16) 4 – 24 121 – 96 64 – 64 –20 25 b2 – 4ac is negative. There are no real solutions b2 – 4ac is positive. There are two real solutions b2 – 4ac is zero. There is one real solution

Find the number of solutions of each equation using the discdriminant. Check It Out! Example 3 Find the number of solutions of each equation using the discdriminant. a. c. b. 2x2 – 2x + 3 = 0 x2 + 4x + 4 = 0 x2 – 9x + 4 = 0 a = 2, b = –2, c = 3 a = 1, b = 4, c = 4 a = 1, b = –9 , c = 4 b2 – 4ac b2 – 4ac b2 – 4ac (–2)2 – 4(2)(3) 42 – 4(1)(4) (–9)2 – 4(1)(4) 4 – 24 16 – 16 81 – 16 –20 65 b2 – 4ac is negative. There are no real solutions b2 – 4ac is zero. There is one real solution b2 – 4ac is positive. There are two real solutions

Assignment Study Guide 10-4 Skills Practice 10-4 Supplemental Worksheet