1. Using the Quadratic Formula to Find Solutions
Section 9.3
Using the Quadratic Formula to Find Solutions

2. Quadratic Formula
The quadratic formula can be used to solve any quadratic equation.
Quadratic Formula
The roots of any quadratic equation of the form
ax2 + bx + c = 0, where a, b, and c are real numbers and a  0, are 2

3. Example Solve using the quadratic formula. 2x2 = 8x + 90
Write in standard form.
Write the quadratic formula and substitute values for a, b, and c.
Simplify.
Continued 3

4. Example (cont) Solve using the quadratic formula. 2x2 = 8x + 90
Solve for x.
Check:
2x2 = 8x + 90
2x2 = 8x + 90
2(9)2 = 8(9) + 90 ?
2(5)2 = 8(5) + 90 ?
162 = ?
50 =  ?
162 = 162 
50 = 50  4

5. Example Find the roots of x2 + 3x – 7 = 0.
Approximate to the nearest thousandth.
a = 1, b = 3, c = –7
Square roots can be found using a calculator or a square root table. 5

6. Example Solve. 3x2 – 4x + 2 = 0 a = 3, b = – 4, c = 2
There is no real number that is
The expression under the radical is the discriminant. 6

7. Discriminant
We can tell whether the roots of any given quadratic equation are real numbers.
Look at the quadratic formula:
1. If the discriminant is a negative number, the roots are not real numbers, and there is no real number solution (no real roots) to the equation.
2. If the discriminant is a positive number, the roots are real numbers, and there are two real number solutions (two real roots) to the equation.
3. If the discriminant is equal to 0, there is one real number solution (one root) to the equation. 7

8. Example Determine whether 3x2 = 5x – 4 has real number solution(s).
First we place the equation in standard form. Then we need only check the discriminant.
The discriminant is negative. Thus 3x2 = 5x – 4 has no real number solution(s). 8