Study Guide

# Squares and Square Roots - Solving Radical Equations

A radical equation is an equation that features a variable contained inside a radicand. At least it won't get wet if it rains.

An example of a radical equation is .

The equation is not a radical equation, because the variable doesn't occur inside the radicand. The 5 and 9 are making it wait outside, apparently, and it's getting drenched.

To solve a radical equation that has a single radical on one side and a number on the other, we square both sides to eliminate the radical and then solve the resulting equation. It'll only get more complicated from here, so enjoy it while it lasts.

### Sample Problem

Solve: .

We square both sides to eliminate the radical. This gives us x + 1 = 49, so x = 48.

Check to make sure this works. Plugging x = 48 into the left-hand side of the equation yields , which is totally 7.

Sometimes we need to rearrange the equation a little so that the radical is all by itself, or "isolated" on one side of the equation. Poor little radical.

### Sample Problem

Solve: .

First we need to add 1 to each side to get the radical all by itself on the left-hand side of the equation. Don't feel badly for him; he's a loner anyway. He'll appreciate the solitude: .

Then, we square both sides to eliminate the radical:

5 + x = 16.

Solving this equation gives us x = 11.

We'll push our double-checking agenda again here. Plug x = 11 into the left-hand side of the equation: ...which is, in fact, 3. Looks like our train hasn't come off the tracks. The two sides of the equation agree, so our solution is x = 11.

As we mentioned, some radical equations have no solution and are so obvious about it that we don't need to do any work. We have no problem with that. Subtlety is overrated.

### Sample Problem

Solve: .

The radical is already by itself. It even hung a "Do Not Disturb" sign on its door. We square both sides:

2x + 6 = x2 + 6x + 9

Now we rearrange for a nice quadratic equation, being sure we've got zero off to one side:

0 = x2 + 4x + 3

This equation factors as:

0 = (x + 3)(x + 1)

So our solutions are x = -3 and x = -1.

As always, we'll cover our butts by checking to make sure both of these are correct solutions.

When x = -3, the left-hand side of the original equation is: And the right-hand side of the original equation is:

-3 + 3 = 0

Since the two sides agree, x = -3 is a solution to the original equation.

When x = -1, the left-hand side of the original equation is: And the right-hand side is:

(-1) + 3 = 2

So x = -1 is also a solution to the original equation. Excellent. Butts officially covered.

Be careful: When solving radical equations, make sure to check your answers. Sometimes they won't actually be solutions to the original equation. You don't want egg on your face. Especially if it's been soft-boiled.

### Sample Problem

Solve: We square both sides to find:

4x + 5 = x2

Rearrange for a nice-looking quadratic equation. Wow, has this thing been working out?

0 = x2 – 4x – 5

This factors as:

0 = (x – 5)(x + 1)

So our solutions should be x = 5 and x = -1. Now we check to make sure both of these are solutions to the original equation.

When x = 5, the left-hand side of the original equation is , and the right-hand side of the original equation is also 5, so x = 5 is a solution.

When x = -1, the left-hand side of the original equation is .

However, when x = -1 the right-hand side of the original equation is -1. Uh-oh.

Since the left-hand and right-hand sides of the equation don't agree for x = -1, this isn't a solution. Argh, so close: x = -1 is an extraneous solution. In other words, it's a number we found through legitimate means, but it's not a solution to the original equation.

It's like if you put in an honest day's work and then got paid in Trident gum. You got something for your efforts, but it's not what you wanted, and it won't help you pay for anything. Although at least it does taste exactly like green apple and golden pineapple. Well, that's something.

Our final answer for this problem is the single solution x = 5.

• ### The Pythagorean Theorem

Before we state the Pythagorean Theorem, we'll introduce a right triangle. Reader, right triangle. Right triangle, reader. We have a feeling you two will get along. You have a lot in common. You both have two legs, are most comfortable at 90 degrees, and are always playing some angle. You also have the same taste in music.

A right triangle is a triangle with a right angle in it, as indicated in the image below by the little box in the lower right corner.

The sides of the triangle that form the right angle are called legs, and the long side across from the right angle is the hypotenuse.

We really, really, really hope you like right triangles. We want them to live up to all the hypotenuse. The Pythagorean Theorem states that, for a right triangle with legs of length a and b and a hypotenuse of length c, the following equation is true:

a2 + b2 = c2

After that dizzying quadratic formula, this one isn't bad at all. We can handle a square on each variable. That sounds okay. However, what does that mean in relation to the right triangle? In words, the sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse. It's a simple concept, but a little wordy, so feel free to go back and read it again. And again. It's okay, read it a few dozen times. You'll be using it a lot. Why is the Pythagorean Theorem true? Because Shmoop says it is? Nah, we'll back up our claims with some proof...this time. Here's one way to see why the Theorem makes sense. Take the right triangle shown above, make 3 copies of it, and lay them out like this: Now we have a big square with side length (a + b), and a smaller square inside with side length c. Who knew triangles could make such great squares? You should see their trapezoids.

We can find the area of the big square in two ways. We know, we know: why can't there ever be only one way?

We can find the area of the big square using its side length, or we can add the areas of the four triangles and the smaller square. With the first way, we have: With the second way, we have: Since a2 + 2ab + b2 and 2ab + c2 are each the area of the big square, they must be equal:

a2 + 2ab + b2 = 2ab + c2

Subtracting 2ab from each side, we see that:

a2 + b2 = c2

Which is the statement of the Pythagorean Theorem. If that wasn't like magic, we don't know what is. Top that, David Blaine.

For more proofs of the Pythagorean Theorem, check this out.

Even if you don't read through them, it's sort of fun to check out how many proofs there are...especially if you find it fun to see how obsessively deranged mathematicians can be. Did we legitimately need that many proofs, guys? We believed you the first time.

We use the Pythagorean Theorem to find out how long different sides of a triangle are. Usually we'll be told the lengths of two sides of the triangle, and then be asked to find the length of the third side. Although we don't know why whoever measured those first two sides couldn't have measured the third one while he was at it. It would have been a nice gesture.

### Sample Problem

A right triangle has one leg that's 3 cm long and another leg that's 4 cm long. How long is its hypotenuse?

We use the Pythagorean Theorem to figure this out. Let's call the hypotenuse c. The two legs have lengths 3 and 4, so 32 + 42 = c2.

Simplifying, we see that 25 = c2.

The solutions to this equation are c = ±5, but since c is a length, we'll only take c = 5 cm. No -5 cm long triangles for us.

You see, it's not enough to arrive at an answer; you'll need to think about it logically and see if it makes sense. If it doesn't, say buh-bye and throw it into the garbage disposal. If you were ever to visit a landfill, you'd be amazed by how many triangles with negative lengths are laying around all over the place.

• ### Word Problems

The Pythagorean Theorem is great for solving word problems. It's also good for sparking some scintillating dinner conversation. At least, we think so; no word yet from that dude who ran screaming from the room. When a word problem describes a situation that can be visualized using a triangle, it's likely that the Pythagorean Theorem is lurking somewhere nearby.

Even Darth Vader, a classic lurker, knows this.

### Sample Problem

Juan and Lenor met for lunch. At 1 p.m., they parted ways. Maybe forever, considering how they left things. Juan drove due south at 30 mph and Lenor drove due east at 60 mph. Apparently, she was more upset than he was. At 1:30 p.m., how far away are Juan and Lenor from each other?

This needs a picture, and there's definitely a triangle in here somewhere. Juan drove south, or straight down, and Lenor drove east, or straight out to the right: The question is asking us to find c, and we've been given enough information to find a and b. The two have been traveling for half an hour...hopefully long enough for them to cool down and regret throwing all that furniture.

Lenor has traveled .

Juan has traveled .

After half an hour, the triangle looks like this: We're supposed to find the distance between Lenor and Juan, so we need to find c. We can do that with the Pythagorean Theorem:

(15)2 + (30)2 = c2
225 + 900 = c2
1125 = c2

Now snag the square root: Negative distance doesn't make sense, although they were practically a negative distance away from each other a half hour ago when they were at each other's throats. Therefore, we take only the positive square root and conclude that, at 1:30 p.m., Juan and Lenor are miles apart (about 33.5 miles).

For the problem above, 33.5 miles would probably be an acceptable answer. Who cares if Juan and Lenor are closer to 33.541 miles apart? However, we gave the answer because that's an exact answer. We may not know all the decimal places of the number , but we know that's exactly how far apart they are at 1:30 p.m.

If a problem says to give an exact answer and your answer has radicals in it, leave the radicals there. The exact answer to the question "what is the square root of 2?" is , not 1.414. It may not surprise you to learn that math is something of an exact science, so those exact answers are important.

Note: Since the writing of this example, we at Shmoop are happy to report that Juan and Lenor have gotten back together, talked, and worked things out. Not a single folding chair was hurled in the process.

### Sample Problem

Find the exact area of a right triangle with a hypotenuse of length 20 and one leg of length 7.

We'd better draw a picture, partly because we look for any excuse to stretch our artistic muscles. Use b for the unknown leg, and make that the base of the triangle: We know the area of a right triangle is .

In other words, the area of a right triangle is one-half the product of the lengths of its legs. We know the length of one leg, and if we had the length of the other leg we'd be all set. Thankfully, we can use the Pythagorean Theorem to find the length of that missing piece. Otherwise, this triangle might need to be fitted for a peg leg.

We know that b2 + 72 = 202, so b2 = 351.

Taking the positive square root (since a triangle can't have a negative base), we see that: This is great. If you write that down as your final answer, however, all your work will have been for nothing. We haven't yet answered the actual question in the problem. Glancing back at the problem to remind ourselves what we were supposed to be looking for (oh, right, the area of the triangle), we see that we now have all the pieces we need. We know the height of the triangle is 7 and its base is , so its area is: That's a little nasty looking, but unfortunately there's no nicer way to write this. We're done. We wash our hands of you, ugly word problem.

Be careful: If you've read through the examples in this section and they made complete sense to you, give yourself a pat on the back—or a pat on the knee, if that's more easily accessible. You've gotten off to an excellent start.

Then, ask yourself this: would you know what to do if you were given similar problems and asked to solve them by yourself without looking at any webpages, books, or notes? Aye, there's the rub. Make sure you know how to approach the exercises, and any other problems available in your textbook or from your teacher, before deciding you're ready to stop studying. If the following exercises give you trouble, you can always review the above examples to figure out where you went wrong. They'll always be there for you, like Ross, Rachel, Chandler, Monica, Joey, and Phoebe. Okay, we'll throw Gunther in there, too.

Here's a hint for all of the problems coming up in the exercise section: draw pictures. Bonus hint: the pictures should actually relate to the problem at hand. Art class is next period.

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