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Squares and Square Roots

Squares and Square Roots

How to Solve a Math Problem

There are three steps to solving a math problem.

  1. Figure out what the problem is asking.
  2. Solve the problem.
  3. Check the answer.

If you like drawing pictures, by all means, draw pictures every step of the way; this is where you can let your inner Walt Disney out of the box. We'll draw pictures in every step of this next example.

Sample Problem

The length of the hypotenuse of a right triangle is 1 greater than the length of one leg, and 2 greater than the length of the other leg. What are the lengths of the sides of the triangle?

1. Figure out what the problem is asking.

We know there's a right triangle involved, so we'll translate this problem into pictures. Bonus points if you can do this on an Etch-a-Sketch.

Given some information about the relative lengths of the sides, we need to figure out how long each side is. Once we know all of the lengths, we'll be able to make fun of the shortest one.

2. Solve the problem.

We'll name the hypotenuse x, so we don't have equations with the word "hypotenuse" in them. If we redraw the picture, it looks like this:

Since this is a right triangle, let's see if the Pythagorean Theorem tells us anything useful. We have a hunch it might.

(x – 1)2 + (x – 2)2 = x2

Distribute the terms on the left-hand side of the equation:

(x2 – 2x + 1) + (x2 – 4x + 4) = x2

Simplify a bit to get x2 – 6x + 5 = 0.

This factors as (x – 5)(x – 1) = 0, which has roots at x = 5 and x = 1.

3. Check the answer.

We found the values x = 5 and x = 1. These are both solutions to the equation (x – 1)2 + (x – 2)2 = x2.

When x = 5, we find that (5 – 1)2 + (5 – 2)2 = 25, which is indeed 5 squared. When x = 1, on the other hand, we find that (1 – 1)2 + (1 – 2)2 = 1, which is indeed 1 squared. However, before jotting down our solutions and calling it a day, we also need to consider the original triangle and make sure both of these make sense. Neither is a negative number on its own, but what about when we plug them into the information provided to us in the original problem?

Plugging in x = 5 gives us this triangle:

Which is totally hunky-dory. But check out what happens when we plug in x = 1:

This so-called triangle is an impostor. A side of length 0 and a side of length -1? Who does it think it's trying to fool? We weren't born yesterday. If we were, by the way, we would be 0 years old, and we would've been -1 days old two days ago.

Anyway, it's a good thing we checked our answers against the original problem, because only one of our answers works: x = 5.

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