If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Integrated math 3

### Course: Integrated math 3>Unit 7

Lesson 3: Extraneous solutions

# Extraneous solutions

Extraneous solutions are values that we get when solving equations that aren't really solutions to the equation. In this video, we explain how and why we get extraneous solutions, by understanding the logic behind the process of solving equations.

## Want to join the conversation?

• At , how is one extraneous if the square root of one equals plus or minus one?
• When we take the square root of one, or just the square root of 5x-4 in general, we're assuming it's the principle square root. If we were taking the plus or minus square root of 5x-4, this would work.

Another way of looking at this, if we square the square root of 5x-4, this allows for a negative value for a principle square root, because the negative value is squared to become positive. But the original equation allows only for positive square roots, because its the principle square root.

If this makes things more perplexing, you just need to know that the original equation takes the principle square root, and thus allows for no negative values.
• In the example where x=1, 1 - 2 = -1. Isn't -1 a square root of 1 and therefore there is no extraneous solution?
• Sal is looking at the principal square root. The principal square root of 1 is positive 1, not negative 1. Principal square roots are always a nonnegative number.
• in the question like this:
square root of 4x is equal to x-3
What are all values of x that satisfy the given equation?
I. 1
II. 9

I choose both because square root of 4 is equal to 2 and -2. But the answer was only II. Is that means on the SAT we should never consider extraneous solution as a true solution?
• That's correct. 'Extraneous' means not relevant to the problem, so we don't accept them as solutions.

Generally, the √x notation refers to the principal root of x when it's used in algebraic equations. We want √x to be a genuine function, which means it can't output two numbers on one input. So if you plug 1 into your equation, it becomes
√4=1-3
2=-2
which is untrue.
• How do I know if a solution is extraneous?
• Plug in your solution back into the original equation. If it shows a false meaning (e.g 2=3) or if the value is undefined (n/0), then it's extraneous.

hopefully that helps !
• Where the -4x + 4 come from? Wouldn't it have been x squared + 2 squared.
• When you square multiple terms, you can't just distribute the exponent like you do with multiplication(You can distribute exponents over multiple factors, however). Instead, you would have to properly multiply out the first binomial by an identical one. Let's say that we had (a + b)^2.
(a + b)^2 = (a + b)(a + b)
= a*a + b*a + a*b + b*b
= a^2 + 2*a*b + b^2
Sal uses a shortcut instead of multiplying everything out, as if you have any numbers a and b, the square of their sum will be a^2 + 2ab + b^2.
• At he says sqrt(1) != -1 but i thought sqrt(1) = +1 or -1.
why does he only take the positive root in this scenario when in many other videos he takes the positive and negative root?
• Because he's talking about the principal sqrt which is always positive
• How would you solve this problem? 1 over 50 + 1 over x = 1 over 30
• 1/50 + 1/x = 1/30
1/x = 1/30 - 1/50
1/x = 50/50 * 1/30 - 30/30 * 1/50
1/x = 50/1500 - 30/1500
1/x = (50 - 30)/1500
1/x = 20/1500
x * 1/x = x * 20/1500
1 = x * 20/1500
1500 * 1 = x * 20/1500 * 1500
1500 = x * 20
1500 / 20 = x * 20 / 20
1500/20 = x
x = 75
• At he says "Well, if a was two and b is three" but that is not possible because a=b?
• At , Sal explains that 0 is an example of an x value for which xa is not equal to xb. Is there another number other than 0 for which that would be true?
• What Sal says is that for 𝑥 = 0,
𝑥𝑎 = 𝑥𝑏 is always true, regardless of whether 𝑎 is equal to 𝑏 or not.

There are no other numbers that do this.
For any 𝑥 ≠ 0, we can divide both sides of the equation by 𝑥:
𝑥𝑎 = 𝑥𝑏 ⇒ 𝑥𝑎∕𝑥 = 𝑥𝑏∕𝑥 ⇒ 𝑎 = 𝑏