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Equation. Discriminant. b 2 – 4 ac. ax 2 + bx + c = 0. 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. EXAMPLE 1. Use the discriminant.
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Equation Discriminant b2 – 4ac ax2 + bx + c = 0 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. EXAMPLE 1 Use the discriminant Number of solutions a.x2+ 1= 0 02– 4(1)(1) = –4 No solution b.x2 – 7 = 0 02 – 4(1)(– 7) = 28 Two solutions (–12)2 –4(4)(9) = 0 c.4x2 – 12x + 9 = 0 One solution
EXAMPLE 2 Find the number of solutions Tell whether the equation 3x2 – 7 = 2xhas two solutions, one solution, or no solution. SOLUTION STEP 1 Write the equation in standard form. 3x2 – 7 = 2x Write equation. 3x2 – 2x – 7 = 0 Subtract 2xfrom each side.
ANSWER The discriminant is positive, so the equation has two solutions. EXAMPLE 2 Find the number of solutions STEP 2 Find the value of the discriminant. b2– 4ac= (–2)2– 4(3)(–7) Substitute 3 for a, – 2 for b, and –7 for c. = 88 Simplify.
ANSWER ANSWER ANSWER two solutions no solution one solution for Examples 1 and 2 GUIDED PRACTICE Tell whether the equation has two solutions, one solution, or no solution. 1. x2 + 4x + 3 = 0 2. 2x2 – 5x + 6 = 0 3. – x2 + 2x = 1