1. WARM UP 4
Simplify 62
(-14)2
-92
-4x2, for x = 3

2. WARM UP 3
Simplify 62
(-14)2
-92
-4x2, for x = 3

3. WARM UP 2
Simplify 62
(-14)2
-92
-4x2, for x = 3

4. WARM UP 1
Simplify 62
(-14)2
-92
-4x2, for x = 3

5. WARM UP
Simplify 62
(-14)2
-92
-4x2, for x = 3

6. 9.2 Solving Quadratic Equations by Finding Square Roots
Real-Life Application: When Will I Ever Use This?
SECURITY AND LAW ENFORCEMENT

7. 9.2 Solving Quadratic Equations by Finding Square Roots
Security and law enforcement officials use many different formulas in their jobs. One such formula is used to determine how fast a car was going when an accident occurred.
𝑺= 𝟓.𝟓 𝑫(𝑭 ±𝒇)
S is the estimated speed.
D is the average length of the actual skid marks.
F is the drag factor.
f is the superelevation (uphill or downhill).

8. 9.2 Solving Quadratic Equations by Finding Square Roots
𝑆= 5.5 𝐷(𝐹 ±𝑓)
S is the estimated speed.
D is the average length of the actual skid marks.
F is the drag factor.
f is the superelevation (uphill or downhill).
A car leaves skid marks whose average length is feet. The drag factor is calculated to be The superelevation is

9. 9.2 Solving Quadratic Equations by Finding Square Roots
𝑆= 5.5 𝐷(𝐹 ±𝑓)
S is the estimated speed.
D is the average length of the actual skid marks.
F is the drag factor.
f is the superelevation (uphill or downhill).
A vehicle leaves skid marks whose average length is 156 feet. The drag factor is calculated to be There is no superelevation.

10. 9.2 Solving Quadratic Equations by Finding Square Roots
𝑆= 5.5 𝐷(𝐹 ±𝑓)
S is the estimated speed.
D is the average length of the actual skid marks.
F is the drag factor.
f is the superelevation (uphill or downhill).
A car leaves skid marks of 199 feet, 185 feet, 203 feet and 215 feet. The drag factor is calculated to 0.87 with a superelevation of (Hint: Remember D is an average.)

11. 9.2 Solving Quadratic Equations by Finding Square Roots
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