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Comment on Equations with Square Roots
But the square root of 9 can
Yes, it is true that there
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
From what i understand we're
I'm assuming you're referring
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
https://gre.myprepclub.com/forum
Hi Brent,
In the aforementioned question, if we take -5 as the value of x then x^2 will be +25. Why is then B>A?
Thanks,
Ketan
Question link: https:/
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
I have a question in mind
If we have an equation, for example, X = Y, X = 5, and Y = -5 will not be a solution to this equation.
Now, if we square both sides of this equation, then X = 5 and Y = -5 will become a solution to this equation. So, is squaring both sides of an equation a legal operation given such circumstances?
The question may be extremely naive, but I had to ask.
That's a great question.
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