Lesson: Equations with Square Roots

Yes, it is true that there are two possible values that, when squared, yield a product of 9.

That is: (3)(3) = 9, and (-3)(-3) = 9

However, the square root NOTATION tells us which root to use.

So, √9 asks to find the POSITIVE number that, when squared, yields a product of 9.

Conversely, -√9 asks to find the NEGATIVE number that, when squared, yields a product of 9.

So, √9 = 3, and -√9 = -3

Does that help?

Cheers,
Brent

I'm assuming you're referring to the question that starts at 4:30

That question involves a square root AND a quadratic equation. So, while quadratic equations don't require us to check for extraneous roots, the square root portion DOES require us to so.

Cheers,
Brent

Question link: https://gre.myprepclub.com/forum/which-is-greater-x-or-11920.html

You're correct about all of that.
However, we aren't comparing x² and 4. We're comparing x and 4.
Since x can be -5 or 5, the answer is D.

Does that help?

Cheers,
Brent

That's a great question.

If we square both sides of an equation, it's possible to get unintended results. For this reason, we must check for extraneous roots (as covered in the above video).

Here's an example: √(2x + 1) = x - 1
Square both sides: 2x + 1 = (x - 1)²
Simplify: 2x + 1 = x² - 2x + 1
Rewrite as: x² - 4x = 0
Factor: x(x - 4) = 0
So, the solutions are x = 0 and x = 4

However, since we squared both sides of an equation, we need to check for extraneous roots by testing each solution.

If x = 0, we get: √[2(0) + 1] = 0 - 1
Simplify √1 = -1
Doesn't work. So, x = 0 is an extraneous root.

If x = 4, we get: √[2(4) + 1] = 4 - 1
Simplify √9 = 3
WORKS! So, x = 4 is a valid solution.

So, the original equation has only 1 solution: x = 4.

Cheers,
Brent