* All skills learned lead eventually to the ability to solve equations and simplify the solutions.*

So we need to figure out at which times does h equal 0. And so we have 8t squared minus 10t minus 25 is equal to 0. And we could complete this square here, or we can just apply the quadratic formula, which is derived from completing the square. Well, that's just positive 100, minus 4 times a, which is 8, times c, which is negative 25. Now, we have to remember, we're trying to find a time.

So we're really solving the equation 0 is equal to negative 16t squared plus 20t plus 50. Or if you're comfortable with this on the left hand side, we can put on the left hand side. And so a time, at least in this problem that we're dealing with, we should only think about positive times.

You now have the necessary skills to solve equations of the second degree, which are known as quadratic equations.

Upon completing this section you should be able to: bx c = 0 when a ≠ 0 and a, b, and c are real numbers.

A border of flowers are planted around the garden so that the area of the flowers is exactly one half the area of the field. The equation area of the garden will be Area = (30 – 2x)(40 – 2x) Area needs to be ½ the area of the field so will be 600m2 600 = (30 – 2x)(40 – 2x) 600 = 1200 – 140x – 4x2 4x2 140x – 600 = 0 (divide by 4) x2 35x – 150 = 0 (factor) (x 30)(x 5) = 0 so x = -30 or x= 5 since distance is positive x = 5.

Give students problems that allow for multiple points of entry and let them use lots of time and technology to develop their own methods to solve them--after several days, you will be excited to see all the different methods they come up with!

Let x be the number of price increases of 10 cents Let y be the revenue a) Revenue = Price x Quantity y = (2.00 0.10x)(300 – 25x) b) The zeros will be when 2.00 0.10x = 0 and 300 – 25x = 0 x = -20 and x = 12 The midpoint is x = -4 There the optimal price will be 4 price decreases.

2.00 0.10(-4) = 1.60 The optimal price will be

Give students problems that allow for multiple points of entry and let them use lots of time and technology to develop their own methods to solve them--after several days, you will be excited to see all the different methods they come up with!

Let x be the number of price increases of 10 cents Let y be the revenue a) Revenue = Price x Quantity y = (2.00 0.10x)(300 – 25x) b) The zeros will be when 2.00 0.10x = 0 and 300 – 25x = 0 x = -20 and x = 12 The midpoint is x = -4 There the optimal price will be 4 price decreases.

2.00 0.10(-4) = 1.60 The optimal price will be $1.60 c) The axis of symmetry is x = -4 from above.

Solution: a) To find halfway point set H =10.5 and solve 10.5 =-4.9t 2 21 4.9t 2 = 10.5 10.5 4.9 t 1.46 t Time must be positive, so the diver is halfway down in about 1.5 seconds.

Jon knows that after 2 s, the rocket was 20m in the air. a) Write an equation to model the daily revenue for the restaurant.

||Give students problems that allow for multiple points of entry and let them use lots of time and technology to develop their own methods to solve them--after several days, you will be excited to see all the different methods they come up with!Let x be the number of price increases of 10 cents Let y be the revenue a) Revenue = Price x Quantity y = (2.00 0.10x)(300 – 25x) b) The zeros will be when 2.00 0.10x = 0 and 300 – 25x = 0 x = -20 and x = 12 The midpoint is x = -4 There the optimal price will be 4 price decreases.2.00 0.10(-4) = 1.60 The optimal price will be $1.60 c) The axis of symmetry is x = -4 from above. Solution: a) To find halfway point set H =10.5 and solve 10.5 =-4.9t 2 21 4.9t 2 = 10.5 10.5 4.9 t 1.46 t Time must be positive, so the diver is halfway down in about 1.5 seconds.Jon knows that after 2 s, the rocket was 20m in the air. a) Write an equation to model the daily revenue for the restaurant.So if this is the height, the ground is when the height is equal to 0. So if we apply it, we get t is equal to negative b. So negative negative 10 is going to be positive 10. And this over here, we have a-- let's see if we can simplify this a little bit. And this is equal to 10 plus or minus-- square root of 900 is 30-- over 16. Or t is equal to 10 minus 30, which is negative 20 over 16. So hitting the ground means-- this literally means that h is equal to 0. Plus or minus the square root of negative 10 squared. The negative sign, negative times a negative, these are going to be positive. And so we get time is equal to 10 plus 30 over 16, is 40 over 16, which is the same thing if we divide the numerator and denominator by 4 to simplify it as 10 over-- or actually even better, divide it by 8-- that's 5 over 2. Divide the numerator and the denominator by 4, you get negative 5 over 4. But let's verify that we definitely are at a height of 0 at 5/2 seconds, or t is equal to 5/2. This expression right over here does give us h is equal to 0. We have negative 16 times 5/2 squared plus 20 times 5/2 plus 50. So this is negative 16 times 25/4 plus-- let's see, if we divide 20 by 2, we get 10. So you divide the left hand side by negative 2, you still get a 0. And so this tells us that the only root that should work is 5/2. I think they really just want us to apply the quadratic formula to this modeling situation. The physics, we go into a lot more depth and give you the conceptual understanding on our physics playlist.

.60 c) The axis of symmetry is x = -4 from above.Solution: a) To find halfway point set H =10.5 and solve 10.5 =-4.9t 2 21 4.9t 2 = 10.5 10.5 4.9 t 1.46 t Time must be positive, so the diver is halfway down in about 1.5 seconds.

Jon knows that after 2 s, the rocket was 20m in the air. a) Write an equation to model the daily revenue for the restaurant.

## Comments Solving Problems With Quadratic Functions

## Applications of Quadratic Functions - Monterey Institute for.

Quadratic functions help forecast business profit and loss, plot the course of moving objects. Next is a word problem you may not think is a quadratic equation.…

## Algebra - Quadratic Equations - Part I Practice Problems

Here is a set of practice problems to accompany the Quadratic Equations - Part I section of the Solving Equations and Inequalities chapter of.…

## Quadratic Function Word Problems Superprof

Our learning resources allow you to improve your Maths skills with exercises of Calculus. See our to reinforce your knowledge of Functions.…

## Real World Examples of Quadratic Equations - Math is Fun

Quadratic equations pop up in many real world situations! Here we have collected some examples for you, and solve each using different methods Factoring.…

## Word Problems Involving Quadratic Equations - Algebra Class

Get instant help with word problems that involve quadratic equations.…

## Quadratic Equations Solved Problems and Practice Questions

In this article we cover quadratic equations – definitions, formats, solved problems and sample questions for practice. A quadratic equation is a.…

## Word Problems with Quadratic Functions - SlideShare

Word Problems Involving Quadratic Equations February 14, 2013 Word Problems involving…

## Quadratic Word Problems Projectile Motion - Purplemath

Demonstrates how to solve typical word problems involving quadratics, including projectile motion.…

## Word problem involving a quadratic function Wyzant Ask An Expert

I use this sort of problem in my college classes to have an example for all the different nuances of quadratic function modelling - word problems.…

## QUADRATIC WORD PROBLEMS

ALGEBRA UNIT 11-GRAPHING QUADRATICS. THE GRAPH OF A QUADRATIC FUNCTION DAY 1. •. The Quadratic Equation is written as.…