1. Use the discriminant EXAMPLE 1 Number of solutions Equation ax 2 + bx + c = 0 Discriminant b 2 – 4ac a. x 2 + 1= 0 b. x 2 – 7 = 0 c. 4x 2 – 12x + 9 = 0 0 2 – 4(1)(1) = –4 No solution 0 2 – 4(1)(– 7) = 28 Two solutions (–12) 2 –4(4)(9) = 0 One solution The Number i The equation x 2 + 1= 0 has no real solutions. In a future course, you will learn that the imaginary number i, defined as, and its opposite are solutions.

2. Find the number of solutions EXAMPLE 2 Tell whether the equation 3x 2 – 7 = 2x has two solutions, one solution, or no solution. SOLUTION STEP 1 Write the equation in standard form. 3x 2 – 7 = 2x 3x 2 – 2x – 7 = 0 Write equation. Subtract 2x from each side.

3. Find the number of solutions EXAMPLE 2 STEP 2 Find the value of the discriminant. b 2 – 4ac = (–2) 2 – 4(3)(–7) = 88 Substitute 3 for a, – 2 for b, and –7 for c. Simplify. The discriminant is positive, so the equation has two solutions. ANSWER

4. Tell whether the equation has two solutions, one solution, or no solution. 1. x 2 + 4x + 3 = 0 GUIDED PRACTICE for Examples 1 and 2 ANSWERtwo solutions 2. 2x 2 – 5x + 6 = 0 ANSWERno solution 3. – x 2 + 2x = 1 ANSWERone solution