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### Course: Digital SAT Math>Unit 4

Lesson 9: Radical, rational, and absolute value equations: foundations

# Radical and rational equations — Harder example

Watch Sal work through a harder Radical and rational equations problem.

## Want to join the conversation?

• lol you could've just stopped at m^2-12m+35 and did -b/a for the sum of roots, I solved this in 1 min
• But there can be some roots that does not satisfy
• shouldn't the answer be 5+7=12
• Yes. Well, sort of. The answer is 12. But the question is asking for the sum of the solutions, rather than the equation stating the sum. So, 12 is the correct answer.
• At first sight of such question, I would never think about doing all of that work!
• I have the exact same doubt about the square root of anything being either positive or negative. I'm getting a -7= sq root of 49
• Whoa, whoa, wait! @ he said that the square root of 16 is 4. But couldn't it also be negative 4? Same for the sqrt of 4 @. Couldn't that also be negative 2? How did he know to pick the positive square root?
• Hi! There can't be a negative number in a square root (It is possible but that is a whole different thing). -4 in a square root would be one of the complex numbers.
• why didnt we just square both sides in the beginning? why did we have to move the 3 to the other side first?
• The reason that Sal didn't square whole left side was because his goal in squaring was to get rid of the square root, which gets the equation into a form that we know how to work with. If we try and square the left side, when we expand it we still get a square root in our equation, which is a problem. This is because when you expand (a+b)^2, you get a perfect square trinomial a^2 + 2ab + b^2. The "2ab" term in this case would be 2(3)(sqrt(6m-26)), which in this case is ugly and has a square root in it, which makes things hard.
Generally, you want to eliminate square roots by isolating them on one side of the equation and then squaring both sides.
• couldn't have just used s=-b/a when turned into quadratic instead of making it a bit too long?
• no, because we should check for extraneous solutions.
• How should I know if to start directly adding a square root first, or if I should add the coefficients to both sides first?
• I think you need to isolate the square root, so get rid of any extra coefficients first.☻
(1 vote)
• okay I have a doubt....at Sal writes "-6m" but according to the formula of:
[ (a-b)^2 = a^2 - 2(a)(b) + b^2]
so, when u put (m-3)^2 it should be equal to m^2 - 2(m)(-3) +(-3)^2 which then becomes:
m^2 - (-6m) + 9 , which results in:
m^2 + 6m + 9
and due to this silly mistake, the entire solution has changed... so can someone please explain this?
• One small thing. When we talk about expanding out a binomial squared, you have two ways of thinking about it. Either you say:
(a-b)^2 = a^2 - 2ab + b^2
where b is the positive version of the number
Or:
(a+b)^2 = a^2 + 2ab + b^2
where a and b can be any sign
I like the second one better because with the first one, you can be tempted to double-count the negative. You should get something like this (first option):
a^2 - 2ab + b^2 = m^2 - 2*m*3 + 3^2
= m^2 - 6m + 9
Or this (second option):
a^2 + 2ab + b^2 = m^2 + 2*m*-3 + (-3)^2
= m^2 - 6m + 9
Both thought processes are perfectly fine, but I'd recommend the second one because it's easier not to double count the negative like you were doing.