Algebra Problems Go-Through 2

Community

Algebra

Algebra Problems Go-Through 2

Author: oLahav

Algebra Problems Go-Through: Part 2, Somewhat tougher

As a follow-up to the popular Welcome to Algebra series, here comes a whole new series of lessons, going through algebra sample problems step-by-step for you to follow.

Word problems with given equations- ???

Some problems give you equations to use and ask you to find some information out using these equations. You would think that would make the questions super-easy, but that's not always true. An example can illustrate that:

"The equation d= -10t ^ 2 + 10t + 60 represents the height of a football, d, at given any time t. If the ball was kicked at t=0 , how long was the ball in the air?"

This question saves you from modeling the situation with an equation yourself, but it forces you to figure out how things work on your own. What do we do?

Look at the type of equation your dealing with. You can find this out from the degree of the expression given- here it's a second degree, a quadratic equation.

Picture the situation in relation to the equation. Use coefficients and -/+ signs. Here we can see a downwards sloping quadratic equation that start at 60 inches when t is 0 seconds, and goes on like this:

After you have a picture, figure out what to plug into the equation to find your answer. Here we're looking for a time, so we need to plug in something for d so we can solve the equation. The question asks how long was the ball in the air- well, when d=0 the ball stopped being in the air, so let's plug in that: -10t ^ 2 + 10t + 60=0 .

Finally, once we have the equation, all you need to do is solve. To solve quadratic equations, we simplify factor them into linear ones, like so:
-10 ^ 2+10t+60=0 divide by 10
t -t -6=0 factor
(t-3)(t+2)=0

Now, either t= -2 or t=3 . Clearly, time isn't negative, so the answer is: the ball was in the air for 3 seconds.

Wait, was that a quadratic equation?

Yes, this was a quadratic equation. Solving them is not as hard as it seems- there are two ways to do so and they're both pretty simple. Let's look at the following example: 2n^ 2+2n -544=0 . That 544 doesn't look 2 friendly, so what do we do to solve this?

Way 1:

Factor the quadratic, like we did earlier. We first simplify by dividing both sides by 2, to get n^ 2+ n -272=0 . Now, if we're smart, we notice that this expression equals (n -16)(n+17)=0 . This means that n can be either 16 or - 17. It's easy, but what if we can't see that factoring mechanism? In that case we use our second way.

Way 2:

Using the quadratic equation. For any equation that looks like aX ^2+bX+c=0 , we have $x=\frac{- b +/- \sqrt{b ^ 2- 4ac}}{2a}$ . So let's plug in our values: $n=\frac{- 1 +/- \sqrt{1 ^ 2- 4(1)(- 272)}}{2(1)}=\frac{1 +/- 33}{2}$ . Using the plus, we get 16, and using the minus, we get - 17. Thus, n is either 16 or - 17. See, it's easy- you just need to memorize that formula.

Can we have another word-problem example?

Here's another one: the product of two consecutive positive integers is 272. What is the value of the smaller integer?

Let's translate this into words- call the smaller integer n. Then n+1 is the consecutive integers, and then n(n+1)=272 , which can be written as n ^2+n- 272=0 . And this we're already solved, so n is either 16 or - 17. Which one is it?

The one thing about quadratics, you'll get two answers, and you have to figure out which one answers the word problem. Here, you can just figure it out right away, as the question specifies positive integer, so 16 has to be the answer. In other cases, you have to figure it out on your own, or both answers may be correct.

So you see, math problems aren't so bad!