Algebra Honors ( Periods 6 & 7)

Solving Problems Involving Quadratic Equations 12-6

You can use quadratic equations to solve problems...
We used the examples in the book to start with:

The park commission wants a new rectangular sign with an area of 25 m² for the visitor center. The length of the sign is to be 4 m longer than the width . To the nearest tenth of a meter, what will be the length and the width of the sign?
Always make sure you check before AND after-- to see what the problem is really asking for...
Let x = the width in meters
then x + 4 = the length in meters

Use the formula for the area of a rectangle to write an equation

x(x+4) = 25
solve it
x(x +4) = 25 becomes
x² + 4x = 25
You can use two methods: the quadratic formula would be my second choice since completing the square works easily here

x² + 4x + 4 = 25 + 4
(x + 2)² = 29
x + 2 = ±√29
You can use your calculator, or the table of square roots... or approximately easily using the method taught in class earlier this year
but to the nearest tenth you get
-2 + √29 ≈3.4
-2 - √29 ≈-7.4

Since you can't have a negative root since a negative length has no meaning... you know the width must be about 3.4 meters and therefore the length is 3.4 +4 or approximately 7.4 meters

Problem 2:
The sum of a number and its square is 156. Find the number

Let x = the number
then
x² + x = 156

x² + x - 156 =0
Using the skills you have for factoring
(x +13)(x-12) = 0
so
x = 13 and x = 12
You have two solutions to this question!!

Problem 3:
The altitude of a triangle is 9 cm less than the base. The area is 143 cm²
What are the altitude and base?

Remember the formula for the area of a triangle is A = ½bh

Let b = the length of the base
then the altitude ( the height) is b-9
so
½(b)(b-9) = 143
b² -9b = 286
b² -9b - 286 = 0
(b -22)(b +13) = 0
b = 22 and b= -13
You can't have a negative length so
the base is 22 cm and the altitude is 13 cm


An object that moves through the air and is solely under the influence of gravity is called a projectile. The approximate height (h) in meters of a projectile at t seconds after it begins its flight from the ground with initial upward velocity v₀ is given by the formula
h = -5t² + v₀t
We can find when such a projectile is at ground level (h=0) by solving
0 = -5t² + v₀t.
If a projectile begins its flight at height c, its approximate height at time t is
h = -5t² + v₀t + c .
We can find when it hits the ground by solving h=0 or
0 = -5t² + v₀t +c.

When a projectile is thrown into the air with an initial vertical velocity of r feet per second, its distance (d) in feet above the starting point t seconds after it is thrown is approximately
d = rt – 16t²