Use the discriminant

EXAMPLE 1

Number of solutions

Equationax2 + bx + c = 0

Discriminantb2 – 4ac

a. x2 + 1= 0

b. x2 – 7 = 0

c. 4x2 – 12x + 9 = 0

02 – 4(1)(1) = –4 No solution

02 – 4(1)(– 7) = 28 Two solutions

(–12)2 –4(4)(9) = 0 One solution

The Number i The equation x2 + 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.

Find the number of solutions

EXAMPLE 2

Tell whether the equation 3x2 – 7 = 2x has two solutions, one solution, or no solution.

SOLUTION

STEP 1 Write the equation in standard form.

3x2 – 7 = 2x

3x2 – 2x – 7 = 0

Write equation.

Subtract 2x from each side.

Find the number of solutionsEXAMPLE 2

STEP 2 Find the value of the discriminant.

b2 – 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

Tell whether the equation has two solutions, one

solution, or no solution.1. x2 + 4x + 3 = 0

GUIDED PRACTICE for Examples 1 and 2

ANSWER two solutions

2. 2x2 – 5x + 6 = 0 ANSWER no solution

3. – x2 + 2x = 1 ANSWER one solution